Show that no group can have its automorphism group cyclic of odd order.

I have shown it only if $G$ is cyclic, but I could not do that if $G$ is not cyclic. Can you help?

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    $\begingroup$ physicsforums.com/showthread.php?t=173148 $\endgroup$ – user9413 May 19 '11 at 15:48
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    $\begingroup$ Note that this is not quite true: a group of order $1$ or $2$ has trivial automorphism group, which is cyclic of odd order! But the proof sketched below accounts for this... $\endgroup$ – Pete L. Clark May 19 '11 at 16:04
  • $\begingroup$ you always have inversion $\endgroup$ – yoyo May 19 '11 at 20:48
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    $\begingroup$ @yoyo: Inversion is an anti-automorphism. $\endgroup$ – Cheerful Parsnip May 19 '11 at 21:53
  • $\begingroup$ I think @Peter L. Clark is right. I see the same question on Page 30 of Derek J.S. Robinson's A Course in the Theory of Groups (GTM 80), with ">1" added to the end. $\endgroup$ – ShinyaSakai Aug 3 '11 at 3:41

Here are some hints:

Let $G$ be a group with a cyclic and odd group of automorphisms.

  1. Since $G/Z$ (Z being the center) is a subgroup of $\operatorname{Aut} G$, it will be cyclic. Deduce that $G = Z$, i.e. $G$ is abelian.
  2. Since $G$ is abelian, what can you say about $x\mapsto x^{-1}$ ? Deduce that $G \cong \bigoplus_{i} \mathbb Z/2\mathbb Z$, i.e. $G$ is an elementary abelian $2$-group.
  3. Now you know a lot about $G$, so you may try to find automorphisms of $G$ that will contradict that $\operatorname{Aut} G$ is cyclic.

I can give more hints if you tell where you're stuck.

-- editted: for step 3: For instance: If there are at least 3 factors involved in the direct product $G \cong \bigoplus_i (\mathbb Z/2\mathbb Z)$ then permuting these factors gives rise to an automorphism (for instance $(a,b,c,...)\mapsto (b,a,c,...)$). This implies $S_3$ appears as subgroup of the automorphism-group so it will surely not be cyclic. If there are $2$ factors $G\cong (\mathbb Z/2\mathbb Z)^2$ and it's easy to see that any permutation of the three involutions of this group is an automorphism.

I think there is a more beautiful way to derive the contradiction but I don't see it right now.

-- editted (much later): I just thought of the more beautiful way: If the direct sum has at least two terms, consider the automorphism that switches these terms $(a,b,c,\dots)\mapsto (b,a,c,\dots)$. This is an automorphism of order 2, a contradiction.

  • $\begingroup$ Thanks for the help, you have been enlightening and sorry for my English, I'm Italian. $\endgroup$ – user11116 May 19 '11 at 15:59
  • $\begingroup$ the light is off, can you explain step 3? $\endgroup$ – user11116 May 19 '11 at 16:38
  • $\begingroup$ @user11116: I have included some detail on that step. $\endgroup$ – Myself May 19 '11 at 16:49
  • $\begingroup$ In 2, you should really have a direct sum, not a direct product: otherwise, you could never get a countably infinite elementary abelian $2$-group. Note also that the existence of the isomorphism is essentially equivalent to the existence of a basis for every vector space over $\mathbf{F}_2$, and that requires the Axiom of Choice; see this previous question for a proof that in the absence of AC, such a group could have trivial automorphism group. $\endgroup$ – Arturo Magidin May 19 '11 at 18:08
  • $\begingroup$ @Arturo Magidin: You're right, I was only thinking of the finite case, free from sorrow about choices. Thank you for noticing. I remember reading that other question by the way, it's really something odd . $\endgroup$ – Myself May 19 '11 at 18:14

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