User arturo magidin - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:01:00Z http://mathoverflow.net/feeds/user/3959 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/132247/quotient-groups-of-the-lower-central-series-of-a-free-group/132252#132252 Answer by Arturo Magidin for quotient groups of the lower central series of a free group Arturo Magidin 2013-05-29T18:31:44Z 2013-06-09T20:01:35Z <p>I don't believe so. </p> <p>I will associate my commutators on the left, so that $[a,b,c]=[[a,b],c]$; and use $[x,y]=x^{-1}y^{-1}xy$; I'm not sure it matters, but that's what I'm used to.</p> <p>Say $k=2$; in $F/F_{4}$, the Hall-Witt identity becomes $$[r,s,t][s,t,r][t,r,s]\equiv 1\pmod{F_{4}}.$$</p> <p>Now consider the case where $a_1=[y_2,y_1]$, $b_1=[y_2,x_1]$, $b_2=[y_1,x_1]$. Then $$[a_1,x_1] \equiv [y_2,y_1,x_1] \equiv [y_2,x_1,y_1][y_1,x_1,y_2]^{-1} \equiv [b_1,y_1][b_2,y_2]^{-1}\pmod{F_4}.$$ Now note that $[a_1,x_1]\in A$, but $[b_1,y_1][b_2,y_2]\in B$. </p> <p>Similar problems are likely to arise with other values of $k$.</p> <p><strong>Edited:</strong> It seems your question was different from what I understood; rather, your $A$ is generated by all commutators of the form $[x_i,a]$ with $a\in F_k$ (modulo $F_{k+2}$), and your $B$ by all commutators of the form $[y_j,b]$ with $b\in F_k$ (modulo $F_{k+2}$).</p> <p>$F_{k}/F_{k+1}$ is free abelian, with basis given by the basic commutators of weight $k$ on $x_1,\ldots,x_n,y_1,\ldots,y_m$; by the Witt Formula, the rank is $$M_{n+m}(k)=\frac{1}{k}\sum_{d|k}\mu(d)(n+m)^{k/d}$$ where $\mu$ is the M&ouml;bius function. The rank of $A$ is at most $nM_{n+m}k$, and the rank of $B$ is at most $mM_{n+m}(k)$. But in general they will be much smaller, since commutators of the form $[c,x_j]$ and $[c,y_j]$, where $c$ is a basic commutator of weight $k$, are seldom basic themselves; so you will have a fair amount of "collision". It would take some rather careful analysis with basic commutator collection to figure it out exactly. I would suggest looking at</p> <ul> <li>Ward, James. <em>Basic commutators.</em> Philosoph. Trans. Roy. Soc. London Ser. A <strong>264</strong> (1969), 343-412, MR 0251148</li> </ul> <p>to see if by any chance Ward has already computed these (or for information on techniques for computing it). </p> <hr/> <p><strong>Added.</strong></p> <p>Let's look at the case of $k=2$. $F_2/F_3$ is freely generated by commutators of the forms $[x_j,x_i]$, $1\leq i\lt j\leq n$, $[y_s,y_r]$, $1\leq r\lt s\leq m$, and $[y_t,x_k]$, $1\leq k\leq n$ and $1\leq t\leq m$.</p> <p>On the other hand, $F_3/F_4$ is freely generated by commutators of the forms </p> <ol> <li>$[x_j,x_i,x_k]$ with $1\leq i\lt j\leq n$, $i\leq k\leq n$; </li> <li>$[x_j,x_i,y_t]$ with $1\leq i\lt j \leq n$, $1\leq t\leq m$; </li> <li>$[y_t,x_k,x_i]$ with $1\leq k\leq i\leq n$, $1\leq t\leq m$; </li> <li>$[y_t,x_k,y_v]$ with $1\leq k\leq n$, $1\leq t\leq m$, $1\leq v\leq m$; and </li> <li>$[y_s,y_r,y_t]$ with $1\leq r\lt s\leq m$, $r\leq t\leq m$. </li> </ol> <p>For $A$, we will obtain all commutators of type 1 and all commutators of type 3; no commutators of type 5; when we consider a commutator of the form $[y_s,y_r]$ and take the commutator with $x_i$, we obtain $[y_s,y_r,x_i] \equiv [y_s,x_i,y_r][y_r,x_i,y_s]^{-1}$. So commutators of type 4 are paired off, except we do not get the ones in which $t=v$. Finally, we obtain all commutators of type 2 because when we take $[y_t,x_j]$ for $a$, and take the commutator with $x_i$, we will obtain $[y_t,x_j,x_i]\equiv [y_t,x_i,x_j][x_j,x_i,y_t]^{-1}$; and since $[y_t,x_i,x_j]\in A$, we obtain the commutator $[x_j,x_i,y_t]$. </p> <p>So, what is the rank of $A$? We have $\binom{n}{3}+2\binom{n}{2}$ commutators of type 1; we have $m\binom{n}{2}$ commutators of type 2 and $m\binom{n}{2}+mn$ of type 3; and we have $n\binom{m}{2}$ generators corresponding to commutators of type 4. This gives $$\mathrm{rank}(A) = \binom{n+1}{3} + (2m+1)\binom{n}{2}+n\binom{m+1}{2}.$$</p> <p>As for $B$, the argument is similar: we get no commutators of type 1, all commutators of type 5 (of which there are $\binom{m}{3}+2\binom{m}{2}$); all commutators of type 2 (of which there are $m\binom{n}{2}$); we also get all of type 4, which gives $nm^2$ commutators; as for type 3, these cannot occur from expressions of the form $[b,y_j]$ with $b\in F_2/F_3$. So the rank of $B$ will be $$\mathrm{rank}(B) = \binom{m+1}{3}+\binom{m}{2} + m\binom{n}{2} + nm^2.$$</p> <p>Finally, for the overlap: according to the Witt formula, the rank of $F_3/F_4$ is $$\frac{1}{3}(n+m)\left( (n+m)^2 - 1\right).$$ So this allows you to compute the rank of the overlap.</p> <p>For larger values of $k$ the computations become more difficult, because not all basic commutators are of the form $[c,x_i]$ or $[c,y_j]$. But this should give you an idea of the kind of thing that is going on (and why I don't want to try doing this for $k=3$).</p> http://mathoverflow.net/questions/18496/example-of-the-completion-of-a-noetherian-domain-at-a-prime-that-is-not-a-domain Example of the completion of a noetherian domain at a prime that is not a domain Arturo Magidin 2010-03-17T15:26:34Z 2013-05-13T07:30:26Z <p>Let $R$ be a Noetherian domain, and let $\mathfrak{p}$ be a prime ideal; consider the completion $\hat R_{\mathfrak{p}}$ of $R$ at $\mathfrak{p}$ (the inverse limit of the system of quotients $R/\mathfrak{p}^n$). If $R$ is a PID, it is easy to see that $\hat R_{\mathfrak{p}}$ is a domain. </p> <p>Someone asked in sci.math if $\hat R_{\mathfrak{p}}$ would always be a domain. I thought it would, but looking at Eisenbud's "Commutative Algebra", I found a reference to a theorem of Larfeldt and Lech that says that if $A$ is any finite-dimensional algebra over a field $k$, then there is a Noetherian local integral domain $R$ with maximal ideal $\mathfrak{m}$ such that $\hat{R_{\mathfrak{M}}}\cong A[[x_1,\ldots,x_n]]$ for some $n$; and so this completion will not be a domain if $A$ is not a domain. I would like to know an example directly, if possible.</p> <p>Does someone know an easy example of a noetherian domain $R$ and a prime ideal $\mathfrak{p}$ such that $\hat R_{\mathfrak{p}}$ is not a domain? Thanks in advance.</p> http://mathoverflow.net/questions/125501/measures-of-non-abelian-ness/125502#125502 Answer by Arturo Magidin for Measures of non-abelian-ness Arturo Magidin 2013-03-25T00:46:00Z 2013-03-25T18:26:01Z <p>Of course, one might say that both $Z(G)$ and $[G,G]$, in a sense, "measure" the non-commutativity of $G$. But they are not very good "quantitative" measures. </p> <p>I think what you are aiming at is a notion introduced by Tur&aacute;n and Erd&#337;s (<em>Some problems of a statistical group theory IV</em>, Acta Math. Acad. of Sci. Hung. <strong>19</strong> (1968), 413-435), the "probability that two elements of $G$ commute": $$P(G) = \frac{\left|\{ (x,y)\in G\times G\mid xy=yx\}\right|}{|G|^2}.$$ In fact, $P(G) = k/|G|$, where $k$ is the number of conjugacy classes of $G$. Gustafson proved that if $G$ is nonabelian then $P(G)\leq 5/8$, and extended the notion to compact groups using Haar measure (W. Gustafson, <em>What is the probability that two group elements commute?</em> American Math. Monthly <strong>80</strong> (1973) 1031-1034). MacHale proved that certain values cannot occur: if $P(G)\gt \frac{1}{2}$, then $P(G) = \frac{1}{2} + \left(\frac{1}{2}\right)^{2s+1}$; and $P(G)$ cannot satisfy $\frac{7}{16} \lt P(G) \lt \frac{1}{2}$. Joseph proved that if $G$ is not commutative and $p$ is the smallest prime that divides $|G|$, then $P(G)\leq \frac{p^2+p-1}{p^3}$ (K.S. Joseph, <em>Commutativity in non-abelian groups</em>, PhD thesis, 1969, UCLA). There's been some other work on this.</p> <p>In the case of $S_3$. $|G|=6$, and the set of pairs $(x,y)$ with $xy=yx$ is, as you note, $18$, so the probability that two elements commutes is precisely your "50% nonabelian". </p> <p>Your second notion seems to be that of looking at $G/[G,G]$, which is the "largest" quotient of $G$ which is abelian. </p> <p><strong>Added:</strong> Since I edited to fix the accent on Erd&#337;s, I'll take the opportunity to add some references:</p> <ul> <li>Desmond MacHale, <em>How commutative can a non-commutative group be?</em>, Math. Gazette <strong>58</strong> (1974), 299-202.</li> <li>David J. Rusin, <em>What is the probability that two elements of a finite group commute?</em>, Pacific J. Math <strong>82</strong> (1979), no. 1, 237-247.</li> <li>Robert Guralnick and Geoff Robinson, <em>On the commuting probability in finite groups</em>, J. Algebra <strong>300</strong> (2006), no. 2, 509-528, MR 2228209 (2007g:60011); <em>Addendum</em>, J. Algebra <strong>319</strong> (2008), no. 4, 1822.</li> </ul> http://mathoverflow.net/questions/122553/p-group-with-large-center/122856#122856 Answer by Arturo Magidin for p-group with large center Arturo Magidin 2013-02-25T04:39:07Z 2013-02-25T17:39:35Z <p>This is much less informative than Ralph's excellent answer, but a quick observation is that, for $p\gt 2$, the groups you are looking for are exactly the groups that are isoclinic to the nonabelian groups of order $p^3$. This observation was made by Philip Hall in <em>The classification of prime-power groups</em>, J. Reine Angew. Math. <strong>182</strong> (1940) 130-141; the observation can be found at the bottom of 136.</p> <p>Recall that two groups $G$ an $K$ are <em>isoclinic</em> if and only if:</p> <ol> <li>$G/Z(G)\cong G/Z(K)$; and</li> <li>$[G,G]\cong [K,K]$; and</li> <li>The isomorphisms can be chosen to be compatible; that is, if $\alpha\colon G/Z(G)\cong K/Z(K)$ and $\beta\colon [G,G]\cong[K,K]$, then for all $g,g'\in G$, $$\beta([gZ(G),g'Z(G)]) = [\alpha(gZ(G)),\alpha(g'Z(G))].$$</li> </ol> <p>The central quotient of your groups are of order $p^2$, and since a nontrivial cyclic group cannot be isomorphic to a central quotient, the central quotient is isomorphic to $C_p\times C_p$; the commutator subgroup is contained in the center, and so the group is of class $2$. Since there is a bilinear alternating map from $(G/Z(G))\times (G/Z(G))$ onto $[G,G]$. <strong>Added:</strong> in general, the map $(G/Z(G))\times (G/Z(G))\to[G,G]$ given by $(xZ(G),yZ(G))\mapsto [x,y]$ has image that generates $[G,G]$; in this case, since $G$ is of class $2$, the map is bilinear; if $G/Z(G)$ is generated by $x$ and $y$, then it follows that $[G,G]$ is generated by $[x,x]$, $[x,y]$, $[y,x]$, and $[y,y]$; the first and last are trivial, and third equals the inverse of the second, so $[G,G]$ is cyclic generated by $[x,y]$. It now follows that the map is in fact onto. Since $G/Z(G)$ is of exponent $p$, then so is $[G,G]$ by the bilinearity of the bracket. Since $G$ is nonabelian, it follows that $[G,G]$ is cyclic of order $p$. It is now straightforward to see that a group with center of index $p^2$ is necessarily isoclinic to the nonabelian groups of order $p^3$. Conversely, if $G$ is isoclinic to the nonabelian groups of order $p^3$, then their central quotients must be of order $p^2$, giving the equivalence.</p> <p>An alternative description in the case of $p$-groups, also given by Hall, is that they are precisely the nonabelian $p$-groups that have at least two abelian subgroups of index $p$. </p> http://mathoverflow.net/questions/115727/maximal-subgroups-of-finite-nilpotent-groups/115728#115728 Answer by Arturo Magidin for maximal subgroups of finite nilpotent groups Arturo Magidin 2012-12-07T17:15:06Z 2012-12-07T17:15:06Z <p>A finite nilpotent group is a direct product of its $p$-parts, and maximal subgroups have prime index; so you have at most four primes dividing the order of the group.</p> <p>If $G$ is a $p$-group, then $G/\Phi(G)$ is an elementary abelian $p$-group; if it has order greater than $p^2$, then it has more than $4$ maximal subgroups; and if $p\gt 3$ and $G/\Phi(G)$ has order $p^2$, then you have more than $4$ maximal subgroups. If $G/\Phi(G)$ is cyclic, then $G$ is cyclic and has a unique maximal subgroup. This reduces the problem to $p=2,3$, $G/\Phi(G)$ of order $p^2$, and then to a restricted way in which you can add direct factors.</p> http://mathoverflow.net/questions/115320/prime-divisor-of-finite-group/115335#115335 Answer by Arturo Magidin for Prime divisor of finite group Arturo Magidin 2012-12-03T20:49:47Z 2012-12-03T20:49:47Z <p>As noted, it is unclear what your sums are over.</p> <p>If the $x_i$ are conjugacy class representatives for all conjugacy classes, then $\sum|\mathrm{cl}_G(x_i)| = |G|$, so the question has an affirmative answer by Cauchy's Theorem.</p> <p>More likely is that the $x_i$ are representatives from the conjugacy classes of elements of order $k$; in that case, the answer is "no". Although the example I give in the comments is a $p$-group, it is easy to turn it into an example which is not a $p$-group: take $p=3$, $k=2$, and let $G=C_2\times C_2\times A$, where $A$ is any nontrivial group of order not divisible by $6$ (e.g., $A=C_5$); then $G$ is not a $p$-group, but an element $(a,b,c)$ has order $2$ if and only if $c=1$ and $a$ or $b$ are nontrivial; so $G$ has exactly three elements of order $2$, each of which is its own conjugacy class, so the sum equals $3$. However, $|G|$ is not divisible by $3$, so $G$ does not have any elements of order $3$. </p> http://mathoverflow.net/questions/107182/center-of-p-groups/107207#107207 Answer by Arturo Magidin for Center of p-groups Arturo Magidin 2012-09-14T19:57:18Z 2012-09-15T03:56:51Z <p><strong>Edit.</strong> In fact, any nontrivial abelian $p$-group $A$ can be realized as the center of a $p$-group with index $p^n$ except in the case $n=1$ (if $A$ is trivial, then it cannot be the center of a nontrivial $p$-group). As has been noted, if $N\subseteq Z(G)$ and $G/N$ is cyclic, then $G$ is abelian, so no group can have a center of prime index. For $n=0$, you can just take $A$ itself.</p> <p>For $n\gt 1$, we can use the same trick as the one used by <a href="http://mathoverflow.net/users/6827/konstantin-ardakov" rel="nofollow">Konstantin Ardakov</a> in the comments: take a group $K$ of order $p^{n+1}$ and class $n$ (such groups are called "$p$-groups of maximal class; I'll give an example below). Such a group $K$ has $Z(K)\cong \mathbf{C}_p$, cyclic of order $p$. Let $k$ be a generator of $Z(K)$. Now let $a\in A$ be an element of order $p$, and take the amalgamated direct product $G=(A\times K)/\langle (a,k^{-1})\rangle$. It is easy to verify that $Z(G)\cong A$, and $G/Z(G)\cong K/Z(K)$, and $K/Z(K)$ has ordder $p^n$. </p> <p>Leedham-Green and McKay's <strong>The Structure of Groups of Prime Power Order</strong> (London Math. Soc. Monographs, new series, no. 27), has several examples of $p$-groups of maximal class in Section 3.1. Here are some: for $p=2$ you can take the dihedral, semidihedral, or generalized quaternion groups of order $2^{n+1}$. For odd prime $p$, the analogue of the dihedral group is as follows: let $K_p$ be the $p$th local cyclotomic number field, let $\mathcal{O}$ be its valuation ring, and let $\theta$ be a primitive $p$th root of unity. Let $\mathfrak{p}=(\theta-1)$ be the maximal ideal of $\mathcal{O}$. Then $\mathcal{O}$ is a $C_p$-module, with the generator acting like multiplication by $\theta$. The ideals $\mathfrak{p}^i$ are invariant under the action. We define $\mathbf{E}_{p^n} = (\mathcal{O}/\mathfrak{p}^{n-1})\rtimes \mathbf{C}_p$. This group has maximal class and order $p^{n}$.</p> <p>(Other examples: $\mathbf{C}_p\wr\mathbf{C}_p$ is a $p$-group of maximal class and order $p^{p+1}$. Or let $A$ be an elementary abelian $p$-group of rank $d$, let $M\in\mathrm{GL}(d,p)$ be the matrix that has $1$s in the diagonal and right above the diagonal, and zeros elsewhere. Then $A\rtimes\langle M_d\rangle$ has maximal class if and only if $3\leq d\leq p$). </p> <hr/> <p>On the other hand, you may want simpler groups, say groups $G$ with $Z(G)\cong A$, $[G:Z(G)]$ of order $p^n$, and $G/Z(G)$ abelian.</p> <p>An old paper of R. Baer, <em>Groups with preassigned central and central quotient groups</em>, Trans. Amer. Math. Soc. <strong>44</strong> (1938), no. 3, 387-412, MR1501973, available on-line <a href="http://www.ams.org/journals/tran/1938-044-03/S0002-9947-1938-1501973-3/S0002-9947-1938-1501973-3.pdf" rel="nofollow">here</a>, can be used to determine which abelian $p$ groups can be embedded as the center of a group of class two with index $p^n$ for any $n\gt 1$. The paper considers the problem addressed in the title, and has both an existence and a uniqueness theorem. The existence theorem is restricted to the case in which the central quotient group is a direct sum of cyclic groups, and the uniqueness theorem is further restricted to the case in which the central quotient is finitely generated.</p> <p>Some notation before stating the main existence result: given an abelian group $A$, $r(A)$ denotes minimum cardinality of a maximal linearly independent subset of $A$ (if $A$ is torsion free or of prime exponent, then any maximal linearly independent subset has $r(A)$ elements, but for more general abelian groups this need not be the case). $A_{t}$ denotes the torsion subgroup of $A$, $A[n]$ the subgroup of elements $x$ such that $nx=0$, and $A(p)$ the subgroup of elements such that $p^ix=0$ for some $i\geq 0$ ($p$ a prime, of course). </p> <p>Define $r(A,0)$ to be the rank of $A/A_{t}$, and $r(A,p^i)$ to be the rank of $(p^{i-1}A(p))/(p^iA(p))$</p> <p>The main result of the paper is:</p> <blockquote> <p><strong>Existence Theorem.</strong> If $A$ is an abelian group and $G$ is a direct sum of cyclic groups, then the following conditions are necessary and sufficient for the existence of a group whose center is $A$ and whose central quotient is isomorphic to $G$:</p> <ol> <li>If $G$ contains elements of order $p^i$, then $A$ contains elements of order $p^i$.</li> <li>If $G$ contains elements of infinite order, then $A$ contains elements of infinite order, or the orders of the elements in $G_t$ are not bounded.</li> <li>If the orders of the elements in $G_t$ are bounded, and $r(G,0)$ is a finite positive integer, then $A$ contains elements of infinite order and $1\lt r(G,0)$. </li> <li>If the orders of the elements in $G_t$ are bounded, and $r(G,0)$ is an odd positive integer, then $A$ contains two independent elements of infinite order ($r(A,0)\gt 1$).</li> <li>If $G=G_t$, $G(p)\neq 0$, and the orders of elements in $G(p)$ are bounded, then $G(p)$ contains at least two independent elements of maximum order.</li> <li>If $G=G_t$, and the orders of elements in $G(p)$ are bounded, then if $r(G,p^{i+k})$ is finite with $k\geq 0$ and $r(G,p^i)$ is odd, then $A$ contains two independent elements of order $p^i$ ($r(p^{i-1}A[p])\gt 1$).</li> </ol> </blockquote> <p>So fix $n\gt 1$; if $A$ is an arbitrary nontrivial abelian $p$-group, then let $G$ be the direct sum of $n$ copies of the cyclic group of order $p$. Then 1 is satisfied, 2, 3, and 4 are vacuously true, and $5$ is true. (Note that condition 5 excludes the possibility of index $p$, though that can be derived directly as has been mentioned).</p> <p>Now, since $G(p)=G$, and $pG=0$, for point 6, note that for any $i\gt 1$ we have $r(G,p^i)=0$; and $r(G,p)=n$; if $n$ is even, then 6 is satisfied vacuously, so you can always obtain a group. You can realize it taking an element $x$ of order $p$ in $A$, letting $G$ be the extraspecial $p$-group of order $p^{2n+1}$ with center generated by $c$, and taking the group $A\times G/\langle (x,z^{-1})\rangle$, an amalgamated direct product; same idea as the construction given by Konstantin Ardakov in the comments.</p> <p>If $n$ is odd, however, then condition $6$ requires that $A$ contain at least two independent elements of order $p$.</p> <p>Since a finite abelian group is capable if and only if it is not cyclic and the two largest invariants are equal, if we realize $A$ as the center of a group $B$ with $[B:A]=p^{2k+1}$, then $B$ must be a direct sum of the form $C_{p^{a_1}}\oplus\cdots\oplus C_{p^{a_r}}$ with $r\gt 1$, $a_1\leq a_2\leq\cdots\leq a_r$, and $a_{r-1}=a_r$; so any $G$ we try to use must have at least three cyclic summands, and so picking a different group $G$ to be a central quotient would only put stronger conditions on $A$ (e.g., requiring $A$ to contain at least two independent elements of order $p^i$ for several different $i$).</p> <p>In summary:</p> <p>Let $A$ be an abelian $p$-group, not necessarily finite, and let $n\gt 0$. Then $A$ can be realized as the center of a $p$-group $H$ of class $2$ with $[H:A]=p^n$ if and only if: (i) $A$ is trivial and $n=0$; (ii) $A$ is nontrivial and $n$ is even; or (iii) $A$ is nontrivial, $n\gt 1$ is odd, and $A$ has at least two independent elements of order $p$. </p> <p>And if $n\neq 1$, then any nontrivial abelian $p$-group $A$ can be realized as the center of a group $G$ with $[G:A]=p^n$.</p> http://mathoverflow.net/questions/106139/when-is-this-diagram-of-tensor-powers-an-equalizer/106154#106154 Answer by Arturo Magidin for When is this diagram of tensor powers an equalizer? Arturo Magidin 2012-09-02T03:38:09Z 2012-09-02T21:09:38Z <p>I think what you want is exactly that the dominion (in the sense of Isbell) of $B$ in $A$ be equal to $B$. </p> <p>Recall that if you are in a category of algebras (in the sense of Universal Algebra), and $B\subseteq A$ is a subalgebra of $A$, then the <strong>dominion of $B$ in $A$</strong> (relative to the category of context) is the set $$\{a\in A\mid \forall C\forall f,g\colon A\to C (f|_B=g|_B\implies f(a)=g(a))\}.$$</p> <p>If $x$ lies in the dominion of $B$ in $A$, then the fact that the two embeddings of $A$ into $A\otimes_B A$ agree on $B$ implies that they agree on $x$; that is, $x\otimes 1=1\otimes x$. Conversely, suppose that the two embeddings of $A$ into $A\otimes_B A$ agree on $x$; if $C$ is any commutative ring and $f,g\colon A\to C$ are two maps that agree on $B$, then the universal property of $A\otimes_B A$ guarantees a homomorphism $\Phi\colon A\otimes_B A\to C$ such that $f = \Phi\circ\lambda$ and $g=\Phi\circ\rho$, where $\lambda$ and $\rho$ are the left and right embeddings; then since $\lambda(x)=\rho(x)$, we conclude that $f(x)=g(x)$. Hence any element of the equalizer is in th dominion.</p> <p>A characterization of dominions in the category of commutative rings is given in the Isbell-Mazet-Silver Zigzag Theorem: the dominion of $B$ in $A$ consists precisely of the elements of $A$ that can be written in the form $XYZ$, where $X$ is a row matrix, $Z$ is a column matrix, $Y$ is a square matrix of the appropriate size, $X$, $Y$, and $Z$ have entries in $A$, and $XY$ and $YZ$ have entries in $B$.</p> <p>(See also this <a href="http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like/139#139" rel="nofollow">previous post</a>.)</p> <p>So the inclusion of $B$ is the equalizer of the left and right embeddings if and only if $B$ is equal to its own dominion in $A$. </p> <p>(Essentially, when the category is right-closed, the dominion of $B$ in $A$ is always equal to the equalizer of the two embeddings $A\to A\amalg_{B}A$; in the category of commutative rings the tensor product functions as a binary coproduct, hence the dominion of $B$ in $A$ is the equalizer of the two embeddings $A\to A\otimes_B A$; but this description does not actually answer your question, it just asks it in a different language; so the actual answer you want is contained in the Isbell-Mazet-Silver Zigzag Theorem, when all is said and done.)</p> http://mathoverflow.net/questions/103402/the-direct-sum-of-injective-modules-need-not-be-injective/103405#103405 Answer by Arturo Magidin for the direct sum of injective modules need not be injective Arturo Magidin 2012-07-28T20:46:29Z 2012-07-29T01:41:38Z <p>The standard proof that I am aware of is actually explicit in this regard.</p> <p>As you note, assume that $I_1\subsetneq I_2\subsetneq\cdots$ is an infinite ascending chain of ideals of $R$, let $E(R/I_n)$ be the injective envelope of $R/I_n$ for each $n$, and let $$E=\bigoplus_{n=1}^{\infty}E(R/I_n)$$ be the their direct sum.</p> <p>Let $I=\bigcup\limits_{i=1}^{\infty} I_n$. </p> <p>For each $n$, let $f_n$ be the composition of the embedding $I\hookrightarrow R$ with the canonical map $R\to E(R/I_n)$ (map to the quotient, then embed into the envelope). </p> <p>Since we have a map from $I$ to each $E(R/I_n)$, we obtain a map $f\colon I\to \mathop{\prod}\limits_{n=1}^{\infty}E(R/I_n)$ by the universal property of the product. In fact, the image of $f$ lies in the direct sum, since for every $x\in I$ there exists $n\in\mathbb{N}$ such that $x\in I_m$ for all $m\geq n$, hence the image of $x$ is $0$ in $R/I_n$. So we have a map $I\to E$. </p> <p>I claim that $f$ does not extend to a module homomorphism $\overline{f}\colon R\to E$ (which it would necessarily do if $E$ were injective). Assume to the contrary that we have an extension $\overline{f}\colon R\to E$. Being a module homomorphism with domain the free module of rank $1$, it is completely determined by $\overline{f}(1)$, and so it has the form $\overline{f}(x) = xe$ for all $x\in R$, where $e=\overline{f}(1)\in E$. </p> <p>Now, let $n_0$ be a positive integer such that the $m$th component of $e$ is $0$ for all $m\geq n_0$. Let $x\in I_{n_0}\setminus I_{n_0-1}$. When we map $x$ to $R/I_{n_0}$, we obtain a nonzero element; hence the $n_0$th component of $f(x)$ must be nonzero (since $R/I_{n_0})$ embeds into $E(R/I_{n_0})$). But $f(x) = \overline{f}(x) = xe$, and the $n_0$th component of $e$ is $0$, hence so is that of $f(x)$, a contradiction.</p> <p>The contradiction arises from the assumption that we can extend the map $f\colon I\to E$ to a module homomorphism $R\to E$. Hence no such extension exists, so $E$ is not injective.</p> http://mathoverflow.net/questions/98395/automorphism-group-of-a-p-group-finitely-generated/98398#98398 Answer by Arturo Magidin for Automorphism Group of a p-group (finitely generated) Arturo Magidin 2012-05-30T16:30:40Z 2012-05-30T16:30:40Z <p>For a specific counterexample to the fact that $p^{2n}$ is not an upper bound, take the elementary abelian group of order $p^3$. It has automorphism group of order $(p^3-1)(p^3-p)(p^3-p^2)$ (pick a basis; the first basis vector can go to any nonzero vector; the second to any vector not in the linear span of the first; the third to any vector not in the linear span of the first two images). For $p=3$, this gives $11232$ automorphisms, larger than $(3^3)^2 = 729$. </p> http://mathoverflow.net/questions/90700/where-is-number-theory-used-in-the-rest-of-mathematics/90720#90720 Answer by Arturo Magidin for Where is number theory used in the rest of mathematics? Arturo Magidin 2012-03-09T16:09:20Z 2012-03-09T16:47:09Z <p>There is some interesting, recent work on the PORC conjecture (which is about Group Theory, specifically, finite $p$-groups). In some sense this is treading close to number theory even in its statement, I guess:</p> <p><strong>PORC Conjecture</strong> (Higman) Let $n$ be a fixed positive integer. The number $f(p,n)$ of nonisomorphic $p$-groups of order $p^n$ is given by a polynomial in $p$ whose coefficients depend only on the residue class of $p$ modulo some fixed $N$. </p> <p>(PORC="Polynomial On Residue Classes")</p> <p>The statement itself makes reference to number theory, of course; but the recent work, by du Sautoy and Vaughan-Lee (<a href="http://arxiv.org/pdf/1106.5530.pdf" rel="nofollow">Non-PORC behaviour of a class of descendant $p$-groups</a>, in the arXiv) delves deep into number theory and arithmetic geometry (as does a previous paper by du Sautoy associating the problem of counting nilpotent groups with elliptic curves). </p> http://mathoverflow.net/questions/78787/varieties-of-groups-with-infinite-relatively-free-group-of-rank-2-finite-infinit Varieties of groups with infinite relatively free group of rank 2 finite, infinite in rank 3 Arturo Magidin 2011-10-21T18:22:57Z 2011-10-21T20:01:36Z <p>Does there exist a variety of groups $\mathfrak{V}$ such that the relatively $\mathfrak{V}$-free group of rank 2 is finite, but the relatively $\mathfrak{V}$-group of rank 3 is infinite?</p> <p>(In other varieties of algebras this can occur; for example, in the variety of all lattices, the free lattice of rank 2 is finite, but the free lattice of rank 3 is infinite.)</p> http://mathoverflow.net/questions/74362/non-isomorphic-finite-rings-with-isomorphic-additive-and-multiplicative-structure Non isomorphic finite rings with isomorphic additive and multiplicative structure Arturo Magidin 2011-09-02T13:43:08Z 2011-09-02T23:38:23Z <p>About a year ago, a colleague asked me the following question:</p> <blockquote> <p>Suppose $(R,+,\cdot)$ and $(S,\oplus,\odot)$ are two rings such that $(R,+)$ is isomorphic, as an abelian group, to $(S,\oplus)$, and $(R,\cdot)$ is isomorphic (as a semigroup/monoid) to $(S,\odot)$. Does it follow that $R$ and $S$ are isomorphic as rings?</p> </blockquote> <p>I gave him the following counterexample: take your favorite field $F$, and let $R=F[x]$ and $S=F[x,y]$, the rings of polynomials in one and two (commuting) variables. They are not isomorphic as rings, yet $(R,+)$ and $(S,+)$ are both isomorphic to the direct sum of countably many copies of $F$, and $(R-\{0\},\cdot)$ and $(S-\{0\},\cdot)$ are both isomorphic to the direct product of $F-\{0\}$ and a direct sum of $\aleph_0|F|$ copies of the free monoid in one letter (and we can add a zero to both and maintain the isomorphism).</p> <p>He mentioned this example in a colloquium yesterday, which got me to thinking: </p> <blockquote> <p><strong>Question.</strong> Is there a counterexample with $R$ and $S$ finite?</p> </blockquote> http://mathoverflow.net/questions/67536/any-subgroup-of-f-g-free-group-with-finite-index-contains-a-term-of-lower-centra/67556#67556 Answer by Arturo Magidin for Any subgroup of f.g. free group with finite index contains a term of lower central series? Arturo Magidin 2011-06-12T02:19:43Z 2011-06-12T19:19:44Z <p>The answer is "no" in both cases.</p> <p>The terms of the lower central series of a group are verbal subgroups. If we let $\gamma_c(G)$ denote the $c$th term of the lower central series of $G$, then for any groups $G$ and $K$ and any group homomorphism $\varphi\colon G\to K$, we have $\varphi(\gamma_c(G))=\gamma_c(\varphi(G))\subseteq \gamma_c(K)$; and if $\varphi$ is onto, then $\varphi(\gamma_c(G))$ maps onto $\gamma_c(K)$, so that we have equality.</p> <p>In particular, if $\pi\colon G\to G/N$ is a quotient map, then $\gamma_c(G/N)$ is trivial if and only if $\gamma_c(G)\subseteq N$. That is, the quotient $G/N$ is nilpotent of class at most $c-1$ if and only if the $c$th term of the lower central series of $G$ is contained in $N$.</p> <p>So let $F$ be a free group, and let $N$ be a normal subgroup of $F$. Then $N$ contains $\gamma_c(F)$ if and only if $F/N$ is nilpotent of class at most $c-1$. </p> <p>So, for example, if $F$ is the free group on two generators, then there is a normal subgroup of $F$ such that $F/N\cong S_3$; since $S_3$ is not nilpotent, $N$ does not contain any term of the lower central series of $F$, even though $N$ is of finite index. </p> http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center/65457#65457 Answer by Arturo Magidin for Quotient of subgroups by center. Arturo Magidin 2011-05-19T17:16:37Z 2011-05-19T17:38:20Z <p>The answer to (1) is "yes." As Carnahan notes, you have $H\cap Z_G\subseteq Z_H$. Since $H/(Z_G\cap H)\cong HZ_G/Z_G$ is isomorphic to a subgroup of $G/Z_G$, then, $H/Z_H$ is a quotient of $H/(Z_G\cap H)$, hence isomorphic to a quotient of a subgroup of $G/Z_G$, so $\mathrm{rank}(H/Z_H) \leq \mathrm{rank}(G/Z_G)$. </p> <p>The answer to (2) is "no". Take $G$ to be the relatively free group of class $2$ and rank $k$ (isomorphic to $F_k/(F_k)_3$, where $F_k$ is the free group of rank $k$ and $(F_k)_3$ is the third term of the lower central series of $F_k$). Then $G^{\rm ab}\cong \mathbb{Z}^k$. Let $H=[G,G]$; then $H$ is free abelian of rank $\binom{k}{2}$. Pick $k\gt 3$ to get that the rank of $H/[H,H]$ is greater than the rang of $G/[G,G]$. If you want to exclude the case where $H$ is abelian itself, just add two of the original generators to the commutator subgroup so the abelianization is of rank $2+\binom{k-1}{2}$. </p> <p>For the modified version of (2) (see comments), it is still not true that $\mathrm{rank}(Z_H)\leq \mathrm{rank}(Z_G)$. Take $G$ as above; then $Z_G=[G,G]$ is free abelian of rank $\binom{k}{2}$. Take $H=\langle [G,G], x_1\rangle$, where $x_1$ is one of the free generators of $G$. Then $H$ is free abelian of rank $\binom{k}{2}+1$. </p> http://mathoverflow.net/questions/8609/favorite-popular-math-book/48709#48709 Answer by Arturo Magidin for Favorite popular math book Arturo Magidin 2010-12-09T04:38:21Z 2010-12-09T22:12:55Z <p><strong>Title:</strong> Mathematics and the Imagination</p> <p><strong>Authors:</strong> Edward Kasner and James Newman</p> <p><strong>Short Description:</strong> A number of chapters covering lots of subjects: counting numbers (including the coining of the word "googol"); $\pi$, $i$, and $e$; geometries, plane and "fancy" ("Lobachevsky's Eiffel Towers and Riemann's Holland Tunnels"); puzzles; paradoxes; chance and probability; topology ("rubber sheet geometry"); calculus. Very much in the spirit of Martin Gardner columns, but from before <em>they</em> existed (the book was originally published in 1940; Gardner began writing his columns in 1956). You can see a preview at <a href="http://books.google.com/books?id=Ad8hAx-6m9oC&amp;printsec=frontcover&amp;dq=%22mathematics+and+the+imagination%22&amp;hl=en&amp;src=bmrr&amp;ei=-VwATYPHEMK78gaeq5WVDw&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCcQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow">Google books</a>.</p> <p>Which also leads me to</p> <p><strong>Author:</strong> Martin Gardner</p> <p><strong>Title:</strong> Various</p> <p><strong>Short Description:</strong> A joy to read.</p> http://mathoverflow.net/questions/46192/union-of-two-proper-subgroups/46198#46198 Answer by Arturo Magidin for Union of two proper subgroups Arturo Magidin 2010-11-16T05:59:08Z 2010-11-16T14:39:04Z <p>The question does not match the title... As has been noted, a group $G$ is the union of proper subgroups if and only if $G$ is not cyclic. No group is the union of two proper subgroups (simple exercise often assigned as homework).</p> <p>A more interesting question is: when is a group a union of $n$ proper subgroups, $n\gt 2$, but no fewer?</p> <p><strong>Theorem</strong> (Scorza) A group $G$ is the union of three proper subgroups if and only if $G$ has a quotient isomorphic to $C_2\times C_2$.</p> <p><strong>Theorem</strong> (Cohn) A group $G$ is the union of four proper subgroups and no fewer if and only if $G$ has a quotient isomorphic to $S_3$, or a quotient isomorphic to $C_3\times C_3$. A group $G$ is a union of five proper subgroups but no fewer if and only if it has a quotient isomorphic to $A_5$. A group is a union of six proper subgroups but no fewer if and only if it has a quotient isomorphic to the dihedral group of order $10$, a quotient isomorphic to $C_5\times C_5$, or a quotient isomorphic to $\langle x,y\mid x^5, y^4, x^2yx^{-1}y^{-1}\rangle$. </p> <p><strong>Theorem</strong> (Tomkinson) There are no groups that are the union of seven proper subgroups but no fewer.</p> <p>I seem to remember it has been shown that for any $n\gt 2$, there is a finite set of groups $S(n)$ so that $G$ is the union of $n$ proper subgroups but no fewer if and only if $G$ has a quotient isomorphic to a group in $S(n)$. The minimal number of subgroups that cover the symmetric and alternating groups $S_k$ and $A_k$ have only been found for smallish values of $k$, though upper and lower bounds are known.</p> http://mathoverflow.net/questions/44605/unions-of-sets-exist/44611#44611 Answer by Arturo Magidin for Unions of sets exist? Arturo Magidin 2010-11-02T20:12:28Z 2010-11-03T06:27:18Z <p>A family of sets is really a set whose elements are sets. In ZFC, the Axiom of Union states (taken from Jech's <strong>Set Theory</strong>):</p> <blockquote> <p>Axiom of Union. For any $X$ there exists a $Y = \bigcup X$.</p> </blockquote> <p>That is: $$\forall X\ \exists Y\ \forall u\ \left( u\in Y \leftrightarrow \exists z(z\in X\wedge u\in z)\right).$$</p> <p>So if you have a <em>family</em> of sets, this will play the role of $X$ in the axiom; the sets in the family are the sets $z$. Thus, you cannot have the cardinalities "grow so fast" that the union will not be a set; that can only occur if your collection of sets is not itself a set but a proper class to begin with (e.g., if you tried to take the "union" of the collection of all ordinals), whether the family is disjoint or not.</p> <p>So now suppose you have a family $X$ of sets, and you want to consider "the" <em>disjoint</em> union of the elements of $X$ (up to a bijection, which are the isomorphisms in the category of sets). Using the Axiom Schema of Replacement, with the function $\mathbf{F}(x) = x\times\{x\}$, there exists a set $Y = \mathbf{F}[X] = \{\mathbf{F}(x)\mid x\in X\}$. Then $Y$ is a set, and the elements of $Y$ are pairwise disjoint: if $\mathbf{F}(x)\cap\mathbf{F}(y)\neq\emptyset$, then there exists $z\in x\times\{x\}$ such that $z\in y\times\{y\}$. So $z=(a,x)$ for some $a\in x$, and $z=(b,y)$ for some $b\in y$; hence $(a,x)=(b,y)$, so $x=y$. So let $Z=\cup Y$. Then $Z$ is a disjoint union of the sets in $X$, and is a set because it is a union of $Y$ (using the axiom of unions), which is itself a set by the Axiom of Replacement.</p> <p><strong>Edit:</strong> Actually, I'm being needlessly complicated in the last paragraph; there is no need to invoke the Axiom of Replacement. Given a family $X$ of sets, we can take $Y=(\cup X)\times X$ (which is a set by the Axioms of Union and Power sets, and the Axiom Schema of Separation) and then take $Z=\{ (a,b)\in Y\mid a\in b\}$. This set achieves the same objective, since for each $x\in X$, the subset $Z_x = \{(a,b)\in Z\mid b=x\}$ is bijectable with $x$, the set $Z$ is the union of the $Z_x$, and $Z_x\cap Z_y\neq \emptyset$ if and only if $x=y$. </p> http://mathoverflow.net/questions/44269/commutator-subgroup-does-not-consist-only-of-commutators/44276#44276 Answer by Arturo Magidin for Commutator subgroup does not consist only of commutators? Arturo Magidin 2010-10-30T20:15:36Z 2010-10-30T20:15:36Z <p>The problem is whether the commutator subgroup may contain elements that are not commutators. One example are the free groups. For instance, in the free group of rank $4$, freely generated by $x$, $y$, $z$, and $w$, the element $[x,y][z,w]$ of the commutator subgroup cannot be written in the form $[a,b]$ for some $a,b$ in the group.</p> <p>The smallest finite examples are groups of order 96; there's two of them, nonisomorphic to each other. (This was a result in Robert Guralnick's thesis). See <a href="http://math.stackexchange.com/questions/7811/derived-subgroups-and-commutators" rel="nofollow">this Math Stack exchange question</a> for a description of these groups, and some references.</p> http://mathoverflow.net/questions/43889/proof-synopsis-collection/44263#44263 Answer by Arturo Magidin for Proof synopsis collection Arturo Magidin 2010-10-30T18:44:31Z 2010-10-30T18:44:31Z <p><strong>Mean Value Theorem:</strong> Tilt your head and apply Rolle's Theorem.</p> http://mathoverflow.net/questions/43820/extremely-messy-proofs/43831#43831 Answer by Arturo Magidin for Extremely messy proofs Arturo Magidin 2010-10-27T16:41:03Z 2010-10-27T21:07:29Z <p>Not from measure theory, alas, but the example that jumps to my mind is Gauss's first proof of Quadratic Reciprocity. It appears in the <em>Disquisitiones Mathematicae</em>. The proof occupies arts. 135 through 144 (five and a half pages in the English edition published by Springer); the proof is by strong induction on $q$ (when $p\lt q$). I don't recall who, but someone once called it a proof by "mathematical revulsion."</p> <p>The proof is quite messy. Gauss argues by cases, considering the congruence classes of $p$ and $q$ modulo $4$, and whether $p$ is or is not a quadratic residue modulo $q$. He actually casts his proof as if it were a proof by minimal counterexample, so he further assumes in some instances that the result does <em>not</em> hold (e.g., for $p\equiv q\equiv 1 \pmod{4}$, either $p$ is a quadratic residue modulo $q$ and $q$ is not one modulo $p$; or $p$ is not a quadratic residue modulo $q$ and $q$ is a quadratic residue modulo $p$). They fall into eight cases, though some of those cases themselves break into subcases. For example, Gauss looks at the case when $p$ and $q$ are both congruent to $1$ modulo $4$, and $\pm p$ is not a residue modulo $q$; then he takes a prime $\ell\neq p$ less than $q$ for which $q$ is not a quadratic residue, and considers the cases in which $\ell\equiv 1 \pmod{4}$ or $\ell\equiv 3 \pmod{4}$ separately; the first subcase itself breaks into four separate <strong>sub</strong>-subcases: since $p\ell$ is a quadratic residue modulo $q$, it is the square of some even $e$; then he considers the case when $e$ is not divisible by either $p$ or $\ell$, when it is divisible by $p$ but not $\ell$; when it is divisible by $\ell$ but not $p$; and when it is divisible by $\ell$ and $p$. And so on. By the time Gauss finally gets to the eighth and final case, he is clearly somewhat exhausted, writing merely "The demonstration is the same as in the preceding case."</p> <p>On the one hand, the proof is pretty much the first proof that one might think to try when encountering the problem. But the different cases are just way too messy, and one quickly loses sight of the forest because one is so intently staring at the beetles in the bark of the tree directly in front. </p> <p>Plenty of other proofs would follow (including five more by Gauss), ranging from the clever to the almost magical (do this, do that, and oops, quadratic reciprocity falls out). </p> http://mathoverflow.net/questions/41744/is-there-a-commutator-theoretic-criterion-for-supersolvability-of-a-group Is there a commutator-theoretic criterion for supersolvability of a group? Arturo Magidin 2010-10-11T02:56:51Z 2010-10-11T03:18:34Z <p>A group $G$ is nilpotent if and only if there is a $c\gt 0$ such that the $(c+1)$st term of the lower central series is trivial. A group $G$ is solvable if and only if there is a $c\gt 0$ such that the $c$th term of the derived series is trivial.</p> <p>Is there some similar criterion for supersolvability, or at least one which is purely commutator-theoretic?</p> http://mathoverflow.net/questions/39490/dualizing-the-definition-of-a-free-group/39501#39501 Answer by Arturo Magidin for Dualizing the definition of a free group Arturo Magidin 2010-09-21T14:19:13Z 2010-09-21T18:13:33Z <p>As has been noted in the comments, your definition of "free group on $S$" is not quite right. The map $g\colon S\to F_S$ is fixed, and is part of the "free group" (that is, the free group on $S$ is the pair $(F_S,g)$, with $g\colon S\to F_S$ a set-theoretic map). The universal property is that for every set map $f\colon S\to G$ into an arbitrary group, there exists a unique homomorphism $\varphi\colon F_S\to G$ such that $g = \varphi f$. But $f$ is not allowed to depend on $g$. </p> <p>It is not hard to see that no such "cofree group" can exist on sets with more than one element. Suppose that you have a set $S$ with more than one element, and a "cofree group" on $S$, $C_S$, together with a set-theoretic map $f\colon C_S\to S$ such that for every group $G$ and every set-theoretic map $g:G\to S$, there exists a unique homomorphism $\varphi\colon G\to C_S$ such that $f = g\varphi$. Let $a\in S$ be different from $f(e)$; then the map $g\colon G\to S$ with $g(x)=a$ cannot factor through $C_S$. </p> <p>As for the case $S=\{s_0\}$, uniqueness of $\varphi$ forces $C_S$ to be the trivial group, because both the zero map and the identity on $C_S$ would satisfy the universal property relative to $f$.</p> <p>The free group construction is the left adjoint of the underlying set functor. In general, left adjoints respect colimits and right adjoints respect limits; that is why the underlying set of a product of groups is the set-theoretic product of the underlying sets (underlying set is the right adjoint, so it respects limits like the product), and why the free group on the disjoint union of two sets is the free product of the free groups on the two sets (disjoint union being the coproduct in $Sets$, free product the coproduct in $Groups$, and coproduct being a colimit). As James Borger notes, if you had a dual of the free group construction, it would be the right adjoint of the underlying set functor and would therefore have to respect colimits. So that the underlying set of a free product of groups would have to be the disjoint union of the underlying sets of the groups. This does not occur, so no such object can exist.</p> <p>P.S. As has also been pointed out in the comments, the universal property is nice and all, and can prove uniqueness, but in general one needs either very high-power categorical/universal algebraic theorems to deduce existence, or one must actually <em>construct</em> the objects in some way. In the case of free groups, while there are many constructions (e.g., as a "big direct product"; see the reference I'm about to give), it is via words or other equivalent constructions (e.g., the fundamental group of a bouqet of circles) that one can get a better handle of them. But if you like universal constructions (nothing wrong with that!) I recommend taking a look at George Bergman's <a href="http://math.berkeley.edu/~gbergman/245/" rel="nofollow">An Invitation to General Algebra and Universal Constructions</a>. It has three different constructions of the free group in Chapter 2.</p> http://mathoverflow.net/questions/38771/dual-schroeder-bernstein-theorem/38833#38833 Answer by Arturo Magidin for Dual Schroeder-Bernstein theorem Arturo Magidin 2010-09-15T16:01:48Z 2010-09-15T16:01:48Z <p>This is only a partial answer because I'm having trouble reconstructing something I <em>think</em> I figured out seven years ago...</p> <p>It would seem the Dual Cantor-Bernstein implies Countable Choice. In a <a href="http://groups.google.com/group/sci.math/msg/28543d2b17d8f4ab" rel="nofollow">post in sci.math</a> in March 2003 discussing the dual of Cantor-Bernstein, Herman Rubin essentially points out that if the dual of Cantor-Bernstein holds, then every infinite set has a denumerable subset; this is equivalent, I believe, to Countable Choice.</p> <p>Let $U$ be an infinite set. Let $A$ be the set of all $n$-tuples of elements of $U$ with $n\gt 0$ and even, and let $B$ be the set of all $n$-tuples of $U$ with $n$ odd. There are surjections from $A$ onto $B$ (delete the first element of the tuple) and from $B$ onto $A$ (for the $1$-tuples, map to a fixed element of $A$; for the rest, delete the first element of the tuple). If we assume the dual of Cantor-Bernstein holds, then there exists a one-to-one function from $f\colon B\to A$ (in fact, a bijection). Rubin writes that "a 1-1 mapping from $B$ to $A$ quickly gives a countable subset of $U$", but right now I'm not quite seeing it...</p> http://mathoverflow.net/questions/38004/slight-extension-to-classification-of-finitely-generated-abelian-groups/38018#38018 Answer by Arturo Magidin for slight extension to classification of finitely generated abelian groups Arturo Magidin 2010-09-07T23:53:52Z 2010-09-08T04:57:52Z <p>I have some general comments.</p> <p><strong>Summary:</strong> such a general theory would be much harder than that for abelian groups (and in fact, would contain the classification of finitely generated abelian groups as a special case), and you would lose a lot of the "good" properties of the category of abelian groups along the way. Many constructions (such as the coproduct) would be different from what we are used to. The advantage is that you can fit into the general context of Universal Algebra, so all of the familiar "abstract nonsense" theory will apply (suitably interpreted). On the other hand, if you restrict to the case you are actually interested in, in which the order of each element divides the order of the distinguished element, then you end up with a much simpler theory, but it falls outside the usual confines of Universal Algebra. You can write down an easy description, but it's not going to be well-behaved relative to homomorphisms.</p> <p><strong>Some details.</strong> To consider the general theory from the point of view of Universal Algebra, you are looking at algebras of type (2,1,0,0), where the binary operation is denoted $+$ (with the usual infix notation), the unary operation is "$-$", one of the zero-ary operations is denoted "$0$", and the second unary operation is denoted "$1$". The operations are required to satisfy the obvious identities: $(a+b)+c = a+(b+c)$, $a+b=b+a$, $a+0=a$, $a+(-a)=0$, and no identity is required of the second unary operation $1$. Homomorphisms are required to respect addition, inversion, map $0$ to $0$, <em>and</em> map $1$ to $1$. The class of abelian groups (as usual) is a variety of algebras of this type, characterized by the identity $1=0$. So you have an embedding from abelian groups to "pointed abelian groups", and this embedding has left adjoint which corresponds to taking the quotient modulo the subgroup generated by $1$. You also have a "forgetful" functor from "pointed abelian groups" to abelian groups, obtained by forgetting about the pointed element; </p> <p>This class, however, is no longer an abelian category. I'm pretty sure of the following, but I could have made some mistakes: The product of two pointed abelian groups $(A,1)$ and $(B,1')$ is, as Henry suggested in the comments, the abelian group $A\times B$ with distinguished element $(1,1')$, and the structure projections are the canonical projections. The free pointed abelian group on a set $X$ will be the direct sum of $|X|+1$ copies of $\mathbb{Z}$, with the generator of the extra copy being the distinguished element. The coproduct of $(A,1)$ and $(B,1')$ will be the quotient of $A\oplus B\oplus\mathbb{Z}$ modulo the subgroup generated by $(1,0,1)$ and $(0,1',1)$, and the distinguished element is the image of $(0,0,1)$. The canonical immersions go are the canonical maps into the direct sum followed by the quotient map.</p> <p>This category includes the entire theory of abelian groups as the class of those groups in which $0=1$; so any classification theory will be much harder than the one for abelian groups. For example, the $1$ generator pointed abelian groups: every $1$-generator pointed abelian group is really a $2$-generator abelian group: it is generated by the given generator <em>plus</em> the distinguished element $1$. So suppose $x$ is the given generator. If $\langle 1\rangle\subseteq\langle x\rangle$, then you get an isomorphism type of the form $(C,C')$, where $C$ is a cyclic group and $C'$ is a subgroup of $C$ ($C$ corresponds to the cyclic group generated by $x$, and $C'$ to the subgroup generated by $1$). But these do not exhaust all possibilities. You also get groups in which $\langle x\rangle\cap\langle 1\rangle = \{0\}$; if the orders are coprime, you are back to the previous case, but if the orders are not coprime then the description is a bit more complicated. And then the case where they intersect nontrivially but neither contains the other, and so on. As you can see, things get complicated even in this simplest of cases...</p> <p>Two pointed abelian groups that are isomorphic as pointed abelian groups will be isomorphic as abelian groups, but you can have to pointed abelian groups whose forgetful image is isomorphic, but that are not isomorphic as pointed abelian groups; for example, $(\mathbb{Z},1)$ and $(\mathbb{Z},2)$ are not isomorphic as pointed abelian groups.</p> <p>The advantage is that you have the entire weaponry from Universal Algebra at your disposal, as well as two reasonably well-behaved functors to abelian groups: the forgetful functor, and the adjoint of the inclusion of the variety of abelian groups to the variety of pointed abelian groups. The disadvantage is that it is unfamiliar, and more complicated than the usual theory of abelian groups. (Think about how things get more difficult going form abelian groups to not-necessarily-abelian groups).</p> <p><strong>Different take:</strong> For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I <em>think</em> that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1|n_2|\cdots|n_k$, and with distinguished element a <em>specific</em> generator of the largest cyclic factor in the case where $n_k>0$, <strike>and an arbitrary nonzero element of the last cyclic factor if $n_k=0$</strike> and a torsion element plus a nonzero element of the last cyclic factor if $n_k=0$. <em>However</em>, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof in the finite case can be done the usual way; for an abelian $p$-group $A$, if $x$ is an element of maximal order then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$, so you can start by picking the distinguished element; otherwise, consider the $p$-parts and think of them as quotients of the original pointed group, with distinguished element the image of the distinguished element; then put them together in the usual way, and the distinguished element will correspond to a generator of the largest cyclic factor. In the infinite case, it gets a bit more complicated, but it should go through: <strike>take $A/A^{tor}$, with distinguished element the image of the distinguished element of $A$; this is a direct sum of cyclic groups of infinite order as an abelian group, and the subgroup generated by the distinguished element is a submodule of rank $1$. So if I remember my modules over PIDs correctly, there is a basis for $A/A^{tor}$ of the form $x_1,\ldots,x_r$, in which $1$ is a multiple of $x_r$; then deal with the torsion part as you would with a normal finite abelian group.</strike> <strong>Edit:</strong> No, the infinite case does not go through like this; see comment. You can get it so that it will be of the form $(a,(0,\ldots,0,d))$, where $a$ is in the torsion part and $(0,\ldots,0,d)$ is an element of the torsion free part with $d>0$, but you may not be able to "get rid" of the $a$. This is problematical.</p> <p>The advantage is that <em>there's</em> the description of all such groups. There are some obvious properties, such as: you can only map from $(A,1)$ to $(B,1)$ if the order of the distinguished element of $A$ is a multiple of the order of the distinguished element of $B$; the map must send the distinguished generator of the largest cyclic factor of the expression for $A$ to the distinguished generator of the largest cyclic factor of $B$. Among the disadvantages is that it is not a "decomposition" in the sense of expressing the pointed group in terms of "smaller" pointed groups, and so it may not lead to any sensible "simplification", just a "description", which is probably not what you want. A more sensible simplification would require you to identify all "product irreducible" pointed abelian groups, and this might be pretty difficult.</p> http://mathoverflow.net/questions/37944/induction-vs-strong-induction/37998#37998 Answer by Arturo Magidin for Induction vs. Strong Induction Arturo Magidin 2010-09-07T19:16:19Z 2010-09-07T19:23:10Z <p>This is in response to Andrej Bauer's comment on Ricky Derner post, as the answer does not seem to fit in the comment section.</p> <p>Suppose we take out the Induction Schema and replace it with: $$(SI)\quad \forall k\Bigl(\bigl(\forall n(n\lt k\rightarrow \phi(n))\bigr)\longrightarrow \phi(k)\Bigr)\Rightarrow \forall m(\phi(m)).$$</p> <p>Now, consider the theory that has the first four Peano Axioms and (SI), instead of the usual Induction schema. The statement "Every natural number is either $0$ or a successor" is a theorem under the usual Peano Axioms, but is not a theorem under this modified axiomatic system. To see this, take $\omega+\omega$ as a model. It satisfies the first four axioms using the usual ordinal successor function, and it satisfies the "strong induction" schema (SI) as well. However, $\omega$ itself, as an element of $\omega+\omega$, is neither a successor nor $0$. (Think of having two copies of the natural numbers, a "red" copy going first and a "blue" copy going second; then the "blue $0$" is neither $0$ nor a successor; you can apply regular induction to the proposition "$n$ is red", but the set you obtain is not all of your set of numbers). </p> <p>So the two theories are not equivalent. (SI) is a theorem in Peano Arithmetic, but regular induction is not a theorem in the theory that has the first four Peano Axioms and (SI).</p> <p>However, if you take the first four Peano Axioms, you add "Every number is either $0$ or a successor" as a "fourth-and-a-half" axiom, and then you take (SI) instead of the usual Induction Schema, then you can prove the usual Induction Schema as a theorem in this system; so the two systems (usual Peano Axioms, and first four Axioms plus the "fourth-and-a-half" axiom plus (SI)) are equivalent. If you want (SI) to be equivalent to regular induction, you need a bit more than just the first four Peano Axioms.</p> http://mathoverflow.net/questions/36452/explicit-description-of-all-morphisms-between-symmetric-groups/36555#36555 Answer by Arturo Magidin for Explicit description of all morphisms between symmetric groups. Arturo Magidin 2010-08-24T14:17:52Z 2010-08-24T14:17:52Z <p>Bret Benesh and Ben Newton determined all pairs $(m,n)$ such that $S_m$ contains a maximal subgroup isomorphic to $S_n$. They are either $(n+1,n)$ with the obvious inclusion (or mapping $S_5$ into the image of a point stabilizer under the outer automorphism of $S_6$); $(\binom{n}{k},n)$, coming from the action of $S_n$ on the subsets of $k$ elements of $\{1,2,\ldots,n\}$; and $((kr)!/(r!)^k k!, kr)$ with $1\lt k,r$, with $S_{kr}$ acting on the the right cosets of a maximal subgroups of the wreath product $S_k\wr S_r$. This appears in <em>A classification of certain maximal subgroups of symmetric groups</em>, <strong>J. Algebra</strong> 304 (no. 2) pp. 1108-1113, MR2265507.</p> <p>Bret later also determined all pairs $(m,n)$ such that $S_m$ has a maximal subgroup isomorphic to $A_n$; such that $A_m$ has a maximal subgroup isomorphic to $S_n$; and such that $A_m$ has a maximal subgroup isomorphic to $A_n$. This appears in the book <em>Computational Group Theory and the Theory of Groups</em>, Contemporary Mathematics 470 (L-C Kappe, R. F. Morse, and me as editors), AMS 2008; the paper is <em>A classification of certain maximal subgroups of alternating groups</em>, pp. 21-26, MR2478411.</p> <p>As pointed out by Jack, this does exhaust all possible embeddings of $S_n$ into $S_k$ (presumably you are okay with the maps that are not embeddings...)</p> http://mathoverflow.net/questions/30750/nilpotent-group-with-ascending-and-descending-central-series-different/30796#30796 Answer by Arturo Magidin for Nilpotent group with ascending and descending central series different? Arturo Magidin 2010-07-06T18:09:03Z 2010-07-06T18:09:03Z <p>Nobody seems to have answered the second query so far; one important class of groups in which the two series coincide is the class of <em>$p$-groups of maximal class</em>. A $p$-group $G$ of order $p^n$ is of <em>maximal class</em> if the nilpotency class of $G$ is $n-1$. The nonabelian groups of order $p^3$ are an example.</p> <p>If $G$ is a $p$-group of maximal class, then the lower and upper central series coincide; this is essentially for the same reason as they do in the case of groups of order $p^3$, as Noah Snyder comments: there just isn't enough room for the series to differ.</p> <p>If $G$ is of order $p^n$ and class $n-1$, then letting $G_2=[G,G]$ and $G_{i+1}=[G_i,G]$ (this is different from the way it is defined in the question as I write this; here, the group has class exactly $c$ if and only if $G_c\neq 1$ and $G_{c+1}=1$), we have $|G_i:G_{i+1}|=p$ for $i=2,3,\ldots,n-1$; similarly, $|Z_{j+1}(G):Z_j(G)| = p$ for $j=0,1,\ldots,n-2$. Since $G_{n-1}\subseteq Z_{1}(G)$ and both are of order $p$, they are equal; taking the quotient gives a group of maximal class and order $p^{n-1}$, and an inductive argument gives the equality among the rest of the terms.</p> http://mathoverflow.net/questions/11964/strong-induction-without-a-base-case/30448#30448 Answer by Arturo Magidin for Strong induction without a base case Arturo Magidin 2010-07-03T19:02:00Z 2010-07-03T19:02:00Z <p>This may qualify, though there is a special case hidden inside the argument.</p> <p>Consider a simplified game of nim in which there are $n>0$ matchsticks, a player may remove $1$, $2$, or $3$ matchsticks each turn, and the player who takes the last matchstick wins.</p> <p><b>Theorem.</b> The first player has a winning strategy if $n\not\equiv 0\pmod{4}$; the second player has a winning strategy if $n\equiv 0\pmod{4}$.</p> <p><i>Proof.</i> Strong induction on $n$. Assume the result holds for all $k\lt n$. If $n\equiv 0\pmod{4}$, then after player 1's turn there will be $k\lt n$ matchsticks left, with $k\not\equiv 0 \pmod{4}$. By the induction hypothesis, the first person to play at this point has a winning strategy, this being player 2; thus, player 2 has a winning strategy.</p> <p>If $n\not\equiv 0\pmod{4}$, then write $n=4\ell + t$ with $1\leq t\leq 3$. Have player 1 take $t$ matchsticks, leaving $4\ell$ matchsticks. If $\ell=0$, player 1 just won. If $\ell>0$, then there are $k\lt n$ matchsticks left, with $k\equiv 0\pmod{4}$. By the induction hypothesis, the player who moves second has a winning strategy, this being the original player 1. So player 1 has a winning strategy in this case. </p> http://mathoverflow.net/questions/30066/cardinality-of-maximal-linearly-independent-subset/30070#30070 Answer by Arturo Magidin for Cardinality of maximal linearly independent subset Arturo Magidin 2010-06-30T16:27:08Z 2010-06-30T16:45:46Z <p>If you consider rings that are not necessarily commutative, here's an example: let $V$ be a countable dimensional vector space over a field $F$, and let $A$ be the ring of all endomorphisms of $A$. I claim that $A\cong A\oplus A$ (as left $A$-modules); if so, then using (and iterating) this isomorphism you can find maximal linearly independent subsets of any finite cardinality. </p> <p>To see that $A\cong A\oplus A$, it suffices to exhibit a two-element $A$-basis for $A$. Let $e_1,e_2,\ldots$ be a basis for $V$. Let $f_1\in A$ be the endomorphism that maps $e_2,e_4,e_6,\ldots$ to $e_1,e_2,e_3,\ldots$, respectively, and maps every odd-indexed basis element to $0$; let $f_2\in A$ be the endomorphism that maps $e_1,e_3,e_5,\ldots$ to $e_1,e_2,e_3,\ldots$, and maps the even-indexed basis elements to $0$. Then $f_1,f_2$ spans $A$: if $\varphi \in A$, then we can write $\varphi$ as $\varphi=gf_1+hf_2$, where $g(e_i)=\varphi(e_{2i})$ and $h(e_j)=\varphi(e_{2j-1})$. To see that $f_1$ and $f_2$ are $A$-linearly independent, suppose that $af_1+bf_2=0$; evaluating at the odd indexed $e_i$ shows that $b(e_j)=0$ for all $j$, and evaluating at the even indexed $e_i$ shows $a(e_j)=0$ for all $j$. Thus, $f_1,f_2$ is also a basis for $A$, which gives an isomorphism $A\cong A\oplus A$. Being bases, they are certainly maximal linearly independent sets. </p> http://mathoverflow.net/questions/22299/what-are-some-examples-of-colorful-language-in-serious-mathematics-papers/50612#50612 Comment by Arturo Magidin Arturo Magidin 2013-06-11T16:51:10Z 2013-06-11T16:51:10Z A bit delayed (just two and a half years...) I'm amazed that webpage still exists; I haven't worked at UNAM for over 10 years. My current webpage includes a few more quotes gathered in 2006. <a href="http://www.ucs.louisiana.edu/~avm1260/lenstra.html" rel="nofollow">ucs.louisiana.edu/~avm1260/lenstra.html</a> http://mathoverflow.net/questions/133321/presentation-of-mathrmsp4-mathbbz Comment by Arturo Magidin Arturo Magidin 2013-06-10T18:24:09Z 2013-06-10T18:24:09Z Which &quot;underlined words&quot;? http://mathoverflow.net/questions/132247/quotient-groups-of-the-lower-central-series-of-a-free-group/132252#132252 Comment by Arturo Magidin Arturo Magidin 2013-06-09T19:57:58Z 2013-06-09T19:57:58Z @qkqh: I'm wrong in type $4$ (there's no factor of $\frac{1}{2}$, but how do you get $2n\binom{m}{2}$ commutators of type 4? You choose two indices between $1$ and $m$ for the $y$s, which gives $\binom{m}{2}$; then you choose the index for $x$, which has $n$ possibilities. For this choice, we get the generator $[y_t,x_k,y_s][y_s,x_k,y_t]^{-1}$. So it's only $n\binom{m}{2}$. And I do not see how you get any commutators of type 3 in $B$ at all. http://mathoverflow.net/questions/132247/quotient-groups-of-the-lower-central-series-of-a-free-group/132252#132252 Comment by Arturo Magidin Arturo Magidin 2013-05-30T14:55:29Z 2013-05-30T14:55:29Z @qkqh: That's lousy notation, then. You are really looking at $\langle [F_k,x_1],\ldots,[F_k,x_n]\rangle\bmod F_{k+2}$, and similarly for $B$. I'll have to think about it. http://mathoverflow.net/questions/131480/what-is-called-this-element Comment by Arturo Magidin Arturo Magidin 2013-05-22T16:55:51Z 2013-05-22T16:55:51Z This is not the correct venue for your questions; mathoverflow is for research-level questions. Perhaps you might consider math.stackexchange.com as a better fit for your questions. http://mathoverflow.net/questions/131014/an-odd-order-finite-group-whose-all-minimal-subgroups-are-unique-is-cyclic-is-it Comment by Arturo Magidin Arturo Magidin 2013-05-18T01:43:44Z 2013-05-18T01:43:44Z (i) Please include your question in the body. (ii) What does it mean &quot;whose all minimal subgroups are unique&quot;? What kind of uniqueness? I would say that each minimal <i>nontrivial</i> subgroup of the Klein 4-group is unique: there are three of them, and none of them is equal to any of the others. So presumably, you have some notion of &quot;unique&quot; at work, but you aren't disclosing it so far. http://mathoverflow.net/questions/130988/why-are-schur-multipliers-of-finite-simple-groups-so-small Comment by Arturo Magidin Arturo Magidin 2013-05-17T18:31:30Z 2013-05-17T18:31:30Z I don't know if some intution may be derived from the following (I'm still struggling to try to understand the Schur multiplier), but there's the following: in &quot;The second homology group of a group; relations among commutators&quot; (Proc. Amer. Math. Soc. 3, (1952). 588–595) C. Miller shows that the second homology/Schur multiplier of $G$ can be interpreted as the group of all relations among formal commutators of elements of $G$, modulo those relations that hold &quot;universally&quot; (i.e., in the free group). I would expect few &quot;nice&quot; relations among commutators in simple groups beyond obvious ones. http://mathoverflow.net/questions/128918/why-didnt-finite-group-theorists-consider-groups-where-all-centralizers-of-non-i Comment by Arturo Magidin Arturo Magidin 2013-05-09T03:36:57Z 2013-05-09T03:36:57Z @solovei: Yes, but &quot;make your title your question&quot; does not mean &quot;don't ask your question anywhere <i>except</i> in the title&quot;, and it also does not mean &quot;start writing in the title, continue in the body as if the title is the first line of your post.&quot; The body of your post should also include the information and the question. http://mathoverflow.net/questions/129717/how-to-compute-the-alexander-polynomial-of-general-torus-knot Comment by Arturo Magidin Arturo Magidin 2013-05-05T22:52:04Z 2013-05-05T22:52:04Z Crossposted to math.SE: <a href="http://math.stackexchange.com/questions/381319/alexanderpolynomial-of-torus-knot" rel="nofollow" title="alexanderpolynomial of torus knot">math.stackexchange.com/questions/381319/&hellip;</a> http://mathoverflow.net/questions/128918/why-didnt-finite-group-theorists-consider-groups-where-all-centralizers-of-non-i Comment by Arturo Magidin Arturo Magidin 2013-04-27T21:12:14Z 2013-04-27T21:12:14Z Do you have a question? Please put it in the body of your post. Do you believe that a book begins at the title on the spine, or on the first page? http://mathoverflow.net/questions/128865/can-one-characterize-amenable-groups-with-c-gx-cyclic-for-all-x-neq-1 Comment by Arturo Magidin Arturo Magidin 2013-04-27T05:18:17Z 2013-04-27T05:18:17Z Am I the only one who is bugged by questions that start in the title instead of being self-contained in the body? http://mathoverflow.net/questions/128151/vectors-and-dimension-of-subspace Comment by Arturo Magidin Arturo Magidin 2013-04-20T03:00:03Z 2013-04-20T03:00:03Z This is not a question of appropriate level for this website. For basic linear algebra questions such as these (particularly, for HW-like problems) please use math.stackexchange. http://mathoverflow.net/questions/127289/homology-groups-of-divisible-and-powered-nilpotent-groups Comment by Arturo Magidin Arturo Magidin 2013-04-11T21:28:33Z 2013-04-11T21:28:33Z In part (2), do you really mean to say &quot;$G$ is a $\pi$-powered nilpotent group&quot;, given that you final note is that &quot;for $G$ nilpotent...&quot;? That is, was (2) supposed to just say &quot;$G$ is a $\pi$-powered group&quot;, without the assumption that it is nilpotent? http://mathoverflow.net/questions/125501/measures-of-non-abelian-ness/125654#125654 Comment by Arturo Magidin Arturo Magidin 2013-03-26T20:14:06Z 2013-03-26T20:14:06Z The link is broken due to the HTML content; should be www-rohan.sdsu.edu/~vadim/ps.pdf http://mathoverflow.net/questions/125527/does-the-poset-of-free-factors-of-a-free-group-form-a-lattice Comment by Arturo Magidin Arturo Magidin 2013-03-25T21:51:05Z 2013-03-25T21:51:05Z The intersection of any finite family of free factors in a free group is again a free factor, though there are infinite families for which this does not hold (see <i>On the intersections of free factors of a free group</i>, by Burns, Chau, and Solitar, Proc. Amer. Math. Soc. <b>64</b> (1977) no 1, 43-44. Also, the intersection of two retracts of a free group is a retract (Bergman, G.M., <i>Supports of derivations, free factorizations, and ranks of fixed subgroups in free groups</i>, Trans. Amer. Math. Soc. <b>351</b> (1999) no 4, 1531-1550. As has been mentioned, join seems harder.