Center of p-groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:38:10Z http://mathoverflow.net/feeds/question/107182 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107182/center-of-p-groups Center of p-groups solovei 2012-09-14T13:53:50Z 2012-09-15T03:56:51Z <p>Can one show that any abelian $p$-group (not necessarily finite) is the center of a $p$-group and of index $p$?</p> http://mathoverflow.net/questions/107182/center-of-p-groups/107183#107183 Answer by Charles Matthews for Center of p-groups Charles Matthews 2012-09-14T14:01:05Z 2012-09-14T14:01:05Z <p>The center of a group cannot be of prime index, because a non-central element must fail to commute with something, which therefore cannot be one of its powers.</p> http://mathoverflow.net/questions/107182/center-of-p-groups/107184#107184 Answer by Konstantin Ardakov for Center of p-groups Konstantin Ardakov 2012-09-14T14:01:33Z 2012-09-14T14:01:33Z <p>If $G$ is a group with centre $Z$ such that $G/Z$ is cyclic, then necessarily $G$ is abelian. So no such group can exist.</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>