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I would appreciate if you could please express your opinion about my proof. I'm not yet very good with automorphisms, so I'm trying to make sure my proofs are OK.

Proof:

Since ${\rm Aut}(G)$ is cyclic, ${\rm Aut}(G)$ is abelian. Thus for any elements $\phi, \psi \in{\rm Aut}(G)$ and some elements $g_i \in G$, $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4=\phi\psi(g_2)\phi\psi(g_1)=\phi\psi(g_2g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian.

The proof seems to be quite straightforward, but I'd rather ask for advice.

Shaun
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sequence
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    Ir cannot be right, there are non-Abelian groups with Abelian automorphism group. About the proof, I do not see a justification for the third equality. Actually the proof does not even use the fact that $\phi$ and $\psi$ commute. – André Nicolas Nov 12 '15 at 06:03
  • For clarity, I could write it this way: $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4$

    But $\phi\psi(g_1)\phi\psi(g_2) = \phi\psi(g_2)\phi\psi(g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian.

    Will this now make a proof?

     – sequence Nov 12 '15 at 22:35
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    Exactly the same comments continue to apply. You did not even use $\phi\psi=\psi\phi$. And anyway $\phi\psi=\psi\phi$ is not enough to prove the original group is Abelian. One needs cyclic. – André Nicolas Nov 12 '15 at 22:56
  • I'm not seeing why my argument, if it shows that $g_3g_4 = g_4g_3$, is not good. – sequence Nov 12 '15 at 23:34
  • But it does not show $g_3g_4=g_4g_3$. In the third equation it essentially asserts, with no justification, that they are equal. What is written makes no progress towards a proof. Tweaking will not do it. – André Nicolas Nov 12 '15 at 23:40
  • I appreciate your comments, but I believe there's something that is not clear still. By definition of a homomorphism, if $\phi\psi(g_1) = g_3$ and $\phi\psi(g_2) = g_4$, then $\phi\psi(g_1)\phi\psi(g_2) = g_3 g_4$. But since $\phi\psi(g_1)\phi\psi(g_2) = \phi\psi(g_2)\phi\psi(g_1)$ and $\phi\psi(g_2)\phi\psi(g_1) = g_4 g_3$ then the result follows. Please let me know why exactly this is not right. – sequence Nov 12 '15 at 23:51
  • What is not right is the assertion that $\phi\psi(g_1)\phi\psi(g_2)=\phi\psi(g_2)\phi\psi(g_1)$. – André Nicolas Nov 12 '15 at 23:56
  • If $Aut(G)$ is cyclic, then any element of it can be expressed as $\phi^k$ for some integer $k$. Now take $\phi^k(g_1g_2) = \phi^r\phi^s(g_1g_2) = \phi^r\phi^s(g_1)\phi^r\phi^s(g_2) = \phi^r\phi^s(g_2)\phi^r\phi^s(g_1)$. Why is this not correct? – sequence Nov 13 '15 at 00:17
  • There is the same unjustified interchange of order in the last equation. As has been suggested in answers, work with the inner automorphisms of the original group. – André Nicolas Nov 13 '15 at 00:21
  • Actually, we could use the property that because $Aut(G)$ is cyclic, $Aut(G)$ must be abelian. Thus for any $\phi \in Aut(G)$, $\phi(g_1g_2)=\phi(g_2g_1) = \phi(g_2)\phi(g_1)=\phi(g_1)\phi(g_2) = g_3g_4=g_4g_3$. Would this be correct? – sequence Nov 13 '15 at 00:37
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    Much earlier I had mentioned that there are non-Abelian groups whose automorphism group is Abelian. So you will never be able to derive Abelianness of $G$ from the Abelianness of its automorphism group. The third equation is unjustified. Every one of your calculations has assumed $G$ is Abelian. – André Nicolas Nov 13 '15 at 00:42

3 Answers3

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There is a nice chain of small results which proves this which continues down the path that Groups suggests. If ${\rm Aut}(G)$ is cyclic, then so is any subgroup of it, in particular ${\rm Inn}(G)$. ${\rm Inn}(G)\cong G/Z(G)$ where $Z(G)$ is the center. If $G/Z(G)$ is cyclic, the group is abelian.

Shaun
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CPM
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  • What says that if $G/Z(G)$ is cyclic then $G$ is abelian. That's what I'm stuck on. – Daniel Donnelly Oct 22 '22 at 10:21
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    @DanielDonnelly https://math.stackexchange.com/questions/63087/if-g-zg-is-cyclic-then-g-is-abelian?noredirect=1&lq=1 has a nice solution to that part – CPM Oct 22 '22 at 16:59
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The $\phi,\psi$ commute, and also the following steps are also OK: $$\phi\psi(g_1g_1)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4.$$ Its not clear in your argument why $g_3g_4=\phi\psi(g_2)\phi\psi(g_1)$?

We are allowed to use commutativity of maps $\phi,\psi$, and we have to conclude commutativity of $g_1,g_2$. You may proceed in following directions.

(1) Consider a specific subgroup of ${\rm Aut}(G)$, namely ${\rm Inn}(G)$. How it is related with $G$?

(2) ${\rm Aut}(G)$ is cyclic, so is ${\rm Inn}(G)$, then using (1), what this will imply?

Shaun
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Groups
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  • For clarity, I could write it this way: $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4$

    But $\phi\psi(g_1)\phi\psi(g_2) = \phi\psi(g_2)\phi\psi(g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian.

    Will this now make a proof?

     – sequence Nov 12 '15 at 22:33
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    why $\phi\psi(g_1)\phi\psi(g_2)=\phi\psi(g_2)\phi\psi(g_1)$? Here, LHS is product of two elements in $G$, and RHS is product of same elements in reverse order; we know automorphisms $\phi,\psi$ commute, but do not know at this stage how elements of $G$ namely $\phi\psi(g_1)$ and $\phi\psi(g_2)$ commute? – Groups Nov 13 '15 at 03:17
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Though I think $Inn(G)$ isomorphic to $G/Z(G)$ is more illustrative of both structure and methods, there is a more “direct” way once you establish $Inn(G)$ (conjugations by elements of G) is cyclic (as $Inn(G) \leq Aut(G)$).

Let $a, b \in G$. Denote conjugation by $g \in G$ by $\psi_g()$. Then because $Inn(G)$ is cyclic, say generated by $\psi_t$ for some $t \in G$, there is some $k$ for which $\psi_a = \psi_t^k$. Similarly, there is some $l$ such that $\psi_b = \psi_t^l$.

Note $\psi_t^m = \psi_{t^m}$. Now, look at $aba^{-1}b^{-1} = \psi_a(b)b^{-1} = \psi_t^k(b)b^{-1} = t^kbt^{-k}b^{-1} = t^k \psi_b(t^{-k}) = t^k\psi_{t^l}(t^{-k}) = t^kt^lt^{-k}t^{-l} = e$

So for all $a, b \in G$ $ab=ba$.

Nimrod Priell
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