Arturo Magidin
|
Registered User
|
|
|
1d |
comment |
What is called this element 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. |
|
May 17 |
comment |
Why are Schur multipliers of finite simple groups so small? 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 "The second homology group of a group; relations among commutators" (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 "universally" (i.e., in the free group). I would expect few "nice" relations among commutators in simple groups beyond obvious ones. |
|
May 9 |
comment |
Why didn’t finite group theorists consider groups where all centralizers of non-identity elements are solvable? @solovei: Yes, but "make your title your question" does not mean "don't ask your question anywhere except in the title", and it also does not mean "start writing in the title, continue in the body as if the title is the first line of your post." The body of your post should also include the information and the question. |
|
May 5 |
comment |
How to compute the Alexander polynomial of general torus knot Crossposted to math.SE: math.stackexchange.com/questions/381319/… |
|
Apr 27 |
comment |
Why didn’t finite group theorists consider groups where all centralizers of non-identity elements are solvable? 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? |
|
Apr 27 |
revised |
Can one characterize amenable groups with $C_G(x)$ cyclic for all $x\neq 1$? make the body self-contained, make title less prone to misreading |
|
Apr 27 |
comment |
Can one characterize amenable groups with $C_G(x)$ cyclic for all $x\neq 1$? Am I the only one who is bugged by questions that start in the title instead of being self-contained in the body? |
|
Apr 11 |
comment |
Homology groups of divisible and powered (nilpotent) groups In part (2), do you really mean to say "$G$ is a $\pi$-powered nilpotent group", given that you final note is that "for $G$ nilpotent..."? That is, was (2) supposed to just say "$G$ is a $\pi$-powered group", without the assumption that it is nilpotent? |
|
Mar 26 |
comment |
Measures of non-abelian-ness The link is broken due to the HTML content; should be www-rohan.sdsu.edu/~vadim/ps.pdf |
|
Mar 25 |
comment |
Does the poset of free factors of a free group form a lattice? 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 On the intersections of free factors of a free group, by Burns, Chau, and Solitar, Proc. Amer. Math. Soc. 64 (1977) no 1, 43-44. Also, the intersection of two retracts of a free group is a retract (Bergman, G.M., Supports of derivations, free factorizations, and ranks of fixed subgroups in free groups, Trans. Amer. Math. Soc. 351 (1999) no 4, 1531-1550. As has been mentioned, join seems harder. |
|
Mar 25 |
awarded | ● Good Answer |
|
Mar 25 |
revised |
Measures of non-abelian-ness fix accent in Erdős, add MacHale, Rusin, Guralnick-Robinson reference |
|
Mar 25 |
awarded | ● Mortarboard |
|
Mar 25 |
awarded | ● Enlightened |
|
Mar 25 |
accepted | Measures of non-abelian-ness |
|
Mar 25 |
awarded | ● Nice Answer |
|
Mar 25 |
comment |
Measures of non-abelian-ness That should be commuting probability, not "commutating probability". Sorry. |
|
Mar 25 |
comment |
Measures of non-abelian-ness On the commutating probability in finite groups, J. Algebra 300 (2006), no. 2, 509-528, MR 2228209 (2007g:60011); Addendum, with sundry references, J. Algebra 319 (2008), no. 4, 1822. Sorry I omitted it from my (brief) list of references. |
|
Mar 25 |
revised |
Measures of non-abelian-ness added 182 characters in body; added 120 characters in body |
|
Mar 25 |
answered | Measures of non-abelian-ness |
|
Mar 18 |
revised |
From reducible polynomial to an irreducible one fix statement |
|
Mar 11 |
revised |
Order of difference of two generators of cyclic group fix display |
|
Mar 1 |
comment |
Annihilator ideals @Ali Taherifar: I thought it might be strictly weaker; still, you may want to take a look at the literature on IN-rings. |
|
Feb 28 |
comment |
Annihilator ideals A ring $R$ is called a "right Ikeda-Nakayama ring" (or right IN-ring) if for any two right ideals $I$ and $J$, $l(I)+l(J)= = l(I\cap J)$. Having a name to add to the property may be useful (though you seem to be asking a bit less than right IN if you require $I$ and $J$ to be two-sided ideals. |
|
Feb 28 |
comment |
Annihilator ideals In the noncommutative setting, if $I$ and $J$ are left/right/two-sided ideals, then $I+J = \{x+y \mid x\in I, y\in J\}$ is a left/right/two-sided ideal. It is certainly a subgroup, and $r(x+y)\in I+J$ if $I$ and $J$ are both left ideals, $(x+y)r\in I+J$ if $I$ and $J$ are right ideals. So I'm not sure what you are going on about... |
|
Feb 28 |
comment |
Classification of generously transitive groups @oeter franek: The problem is that the parser interprets < as an HTML marker; use \lt for < and \gt for >. I've fixed the question. |
|
Feb 28 |
revised |
Classification of generously transitive groups fix |
|
Feb 25 |
revised |
p-group with large center add details in light of comment |
|
Feb 25 |
comment |
p-group with large center @Johannes Hahn: The map has image which generates $[G,G]$; since the bilinear form is alternating, in the case at hand the image is generated by $[x,y]$ (where $G/Z(G)$ is generated by the images of$x$ and $y$), hence $[G,G]$ is cyclic; and since the map is bilinear, it will perforce be onto in this situation (rather than merely mapping onto a generating set). Since $G/Z(G)$ is of exponent $p$, $[x,y]$ is of order $p$, and so $[G,G]$ is cyclic of order $p$. |
|
Feb 25 |
answered | p-group with large center |
|
Feb 21 |
comment |
p-group with large center @Hamid: No, there's many such groups, obtained by varying $A$. E.g., with $n=5$, you can have $A=C_{p}^3$, yielding a group of exponent $p$, or $A=C_{p^3}$, yielding a group with an element of order $p^3$. |
|
Feb 11 |
awarded | ● Yearling |
|
Jan 15 |
comment |
Useless math that became useful @Emil: Thanks; in any case, its use above is incorrect, because we do not want to compare and contrast the statement "Number Theory was considered useless) with Hardy's writings on the subject. Rather, Hardy is a reference for this assertion. It should be "e.g." or "see, e.g.". |
|
Dec 17 |
comment |
Useless math that became useful Pet peeve: "cf" stands for "conferre", which means "to compare"; you are using it as reference or a "see for example". Though an extremely common usage, it is incorrect. "cf" should be used for "compare with", and you don't want to compare the writings of Hardy with the statement that Number Theory was considered useless; rather, you want to use Hardy's writings as a reference to the assertion that Number Theory was considered useless... |
|
Dec 9 |
revised |
finite groups with trivial frattini subgroup use `\langle` and `\rangle` instead of `<` and `>`; better spacing |
|
Dec 8 |
comment |
finite groups with trivial frattini subgroup Note: The OP has changed the question to add the condition that at least one maximal subgroup not have prime order. |
|
Dec 8 |
comment |
finite groups with trivial frattini subgroup @higwain: When you change the question so that one of the answers becomes incorrect, then you should do the change in a way that makes it clear and obvious that you changed the question. Here you have added the condition "at least one of its maximal subgroups isn't of prime order", which makes the answer that had already been posted by majid arezoomand look as if majid didn't read your question carefully enough; you have, in essence, made it so that instead of it looking like an omission on your part, it looks like an error on his part. |
|
Dec 7 |
comment |
Groups that do not exist @36min: The $p^aq^b$ theorem is due to Burnside, not Frobenius... |
|
Dec 7 |
answered | maximal subgroups of finite nilpotent groups |
|
Dec 5 |
comment |
What are the relations between conjugates and commutators? It's probably not quite what you are asking, but I'll mention it anyway: A lot of the relations among commutators (and with conjugation, and especially the interaction of commutators and powers) are considered in the study of "commutator collection" and "basic commutators". The big work on them is Ward's "Basic Commutators", Philos. Trans. Roy. Soc. London Series A, vol 264 (1969), 343-412, MR 0251148 |
|
Dec 4 |
accepted | Prime divisor of finite group |
|
Dec 3 |
answered | Prime divisor of finite group |
|
Dec 3 |
comment |
Order of column vectors in jordan normal form It depends on whether you like your Jordan blocks to have the 1s above the diagonal or below the diagonal. The two forms are similar, though. They should appear in $P$ in whatever order you choose for your ordered basis. In the example in Wikipedia, $x$ is a generalized eigenvector but not an eigenvector, and $y=(A-4I)x$ is an eigenvector. Since in the Jordan form the eigenvector occurs before the generalized eigenvector (the third column corresponds to an eigenvector, but the fourth column does not), then the eigenvector occurs first in $P^{-1}$ (following your ntoation, not Wikipedia's) |
|
Dec 3 |
comment |
Prime divisor of finite group @user123: (i) You have failed to clarify what you mean with yoru ntoation. (ii) Take the direct product of the Klein 4-group and a group of order not divisible by 6. You still get exactly 3 elements of order 2, but no elements of order 3. |
|
Dec 3 |
comment |
Prime divisor of finite group If you are adding the order of all conjugacy classes, then you get the order of $G$, so the result follows from Cauchy's Theorem. If you are only adding conjugacy classes corresponding to elements of a given order, then the answer is no: for the Klein $4$-group, $G$ has 3 elements of order $2$, so $3$ divides the sum of sizes of conjugacy classes of elements of order $2$, but $G$ has no elements of order $3$. |
|
Dec 3 |
comment |
Varieties Of Groups & Enumeration Of Size of Isomorphic Factor Groups Derek: Thanks, for the counterexample as well. |
|
Dec 2 |
comment |
Varieties Of Groups & Enumeration Of Size of Isomorphic Factor Groups (In fact, I was in the middle of trying to write an answer when Derek Holt posted his, and it seemed to match what I was going for, so I stopped. I'm a bit worried that he now says the details were wrong...) |
|
Dec 2 |
comment |
Varieties Of Groups & Enumeration Of Size of Isomorphic Factor Groups @Jeremy: Hmmm... I had envisioned something just like Derek Holt wrote (invoking only the universal property), but he has now withdrawn his answer, so I may have overlooked something. The idea is to try to leverage the maps $F\to F/M$ and $F\to F/N\to F/M$ into a map $F\to F$ that maps $N$ into $M$, and then vice-versa and use uniqueness to deduce the desired result. Let me think about it and see if I can figure out the details... |
