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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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    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. – 2011-08-03

1 Answers 1

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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.

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    The elementary $2$-group $G$ can be written as $\oplus_{i} \mathbb{Z} /2 \mathbb{Z}$, and the number of direct summands is larger than $2$. Thus, $\mathrm{Aut} \mathbb{Z} /2 \mathbb{Z} \oplus \mathbb{Z} /2 \mathbb{Z}$ is a subgroup of $\mathrm{Aut} G$. But $\mathrm{Aut} \mathbb{Z} /2 \mathbb{Z} \oplus \mathbb{Z} /2 \mathbb{Z} \cong GL(2,2)$ and are of order $6$. This also contradicts the fact that $\mathrm{Aut}G$ is of odd order. 【Is this all right?】 – 2011-10-22