Timeline for A double centralizing theorem for finite groups
Current License: CC BY-SA 3.0
23 events
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Jul 12, 2014 at 19:00 | history | edited | Sh.M1972 | CC BY-SA 3.0 |
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Jul 11, 2014 at 0:37 | history | edited | Sh.M1972 | CC BY-SA 3.0 |
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Jul 11, 2014 at 0:34 | vote | accept | Sh.M1972 | ||
Jul 10, 2014 at 21:05 | comment | added | Geoff Robinson | When G is Abelian of order 8 and exponent 4, there is a unique characteristic subgroup of order 2, and this is indeed centralized by Aut(G). But the other two subgroups of order 2 of G are transitively permuted by Aut(G), and taking s as a generator of either of these produces an element s such that $C_{G}(A)$ is non-cyclic, where $A=C_{{\rm Aut}(G)}(s).$ | |
Jul 10, 2014 at 20:50 | vote | accept | Sh.M1972 | ||
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Jul 10, 2014 at 20:48 | comment | added | Khalid Bou-Rabee | @DerekHolt I am sorry for that comment. I was confused. | |
Jul 10, 2014 at 20:40 | history | edited | Sh.M1972 | CC BY-SA 3.0 |
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Jul 10, 2014 at 17:41 | review | Close votes | |||
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Jul 10, 2014 at 15:27 | comment | added | Geoff Robinson | I agree that Derek's example is already a counterexample when $p =2.$ | |
Jul 10, 2014 at 8:44 | comment | added | Derek Holt | @MartyIsaacs Sorry I am completely confused, and I do not understand the comment in Khalid Bou-Rabee's answer that I am using an alternative interpretation of $C_H(A)$. In my example, since not all automorphisms of $G$ fix $s$, how can $A$ be the full automorphism group of $G$? | |
Jul 10, 2014 at 8:26 | history | edited | Sh.M1972 | CC BY-SA 3.0 |
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Jul 10, 2014 at 8:13 | history | edited | Sh.M1972 | CC BY-SA 3.0 |
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Jul 10, 2014 at 7:00 | vote | accept | Sh.M1972 | ||
Jul 10, 2014 at 20:50 | |||||
Jul 10, 2014 at 6:57 | answer | added | Geoff Robinson | timeline score: 8 | |
Jul 10, 2014 at 6:12 | comment | added | Geoff Robinson | Note that this group is just the dihedral group of order $8.$ All of its automorphisms do, of course, fix the unique central involution, so if we take $s$ to be a non-central involution, we seem to obtain a counterexample to the assertion of the problem, as Khalid Bou-Rabee points out. | |
Jul 10, 2014 at 3:57 | answer | added | Khalid Bou-Rabee | timeline score: 10 | |
Jul 10, 2014 at 1:15 | comment | added | Sh.M1972 | @Khalid Bou-Rabee: I don't know about the structure of generalized Heisenberg group. It is good to explain some details; which automorphisms belong to $A$? Why they fix elements of $Z(G)$? | |
Jul 9, 2014 at 23:40 | comment | added | Khalid Bou-Rabee | I'm sorry, I'm still confused. Let $G$ be a higher dimensional (greater than 3) generalized Heisenberg group defined over $F_2$. Select a non-central standard generator, $b$, then if $A = C_{Aut(G)}(b)$ is the set of all automorphisms that fix $b$, then $C_G(A)$, by Marty's definition, is the set of all elements of $G$ that are fixed by $A$, which is not cyclic cause it contains $<b>$ and $Z(G)$ which both generate $F_2 \times F_2$. | |
Jul 9, 2014 at 22:18 | comment | added | Marty Isaacs | @Derek No. In your example, A is the full automorphism group of G so C_G(A) is the set of all elements of G that are fixed by all automorphisms, and this is the cyclic group <t^2>. | |
Jul 9, 2014 at 21:59 | comment | added | Derek Holt | I may also be misunderstanding the statement. If we let $G = \langle t \rangle \times \langle s \rangle$ with $|t|=4$, $|s|=2$, then all elements of ${\rm Aut}(G)$ centralize $t^2$, so $C_G(A)$ contains $\langle t^2,s \rangle$, which is not cyclic. | |
Jul 9, 2014 at 21:59 | comment | added | Johannes Hahn | @KhalidBou-Rabee: This is standard notation. If a group $A$ operates on $G$ through automorphisms, then $C_A(S) := \{a\in A \mid {^a s} = s\}$ | |
Jul 9, 2014 at 20:19 | comment | added | Khalid Bou-Rabee | Please clarify your notation. What do you mean by $C_H(S)$? Is this the centralizer of the set $S$ in $H$? If so, then if $G$ is a commutative group, how do you view elements of it inside $Aut(G)$? | |
Jul 9, 2014 at 20:07 | history | asked | Sh.M1972 | CC BY-SA 3.0 |