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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
Jul 11, 2014 at 0:34
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
Jul 11, 2014 at 10:22
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