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Suppose $n \geqslant 3$ such that $n$ is an odd integer, prove that there does not exist a group $G$ such that $\operatorname{Aut}(G)$ is isomorphic to $\mathbb{Z}_n$.

My idea is that if they are isomorphic, then they have the same number of elements of the same order. I try to find an element of order $2$ in $\operatorname{Aut}(G)$, then I can conclude such $G$ does not exist. However, I cannot find an element of order $2$ for $\operatorname{Aut}(G)$...

justadzr
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Joe
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2 Answers2

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Can you show that any Abelian group of order $\ge3$ has an automorphism of order $2$?

If so, you can reduce to the non-Abelian case. If $G$ is non-Abelian, its inner automorphisms form a group isomorphic to $G/Z$ where $Z$ is its centre. But if $G/Z$ is cyclic, it's easy to prove that $G$ is Abelian, a contradiction.

Angina Seng
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  • Showing that any abelian group of order $\geq3$ has an automorphism of order $2$ is non-trivial and relies on choice (unless $G$ is assumed to be finitely generated, which it is not here). See here for the appropriate working. – user1729 Nov 20 '19 at 09:23
  • (If we do not assume the axiom of choice then it is consistent that there exists a group $G$ of order $\geq3$ such that $\operatorname{Aut}(G)$ is trivial, and hence has no elements of order $2$. This is explained in this answer of Asaf Karagila.) – user1729 Nov 20 '19 at 09:29
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Consider the automorphism $\alpha: x \to x^{-1}$. For any abelian group G it is an automorphism, and whenever $|G| > 2$, it has order $2$ (in $C_2$ and the trivial group it's simply the identity). Hence $\operatorname{Aut}G \not\cong \mathbb{Z}_n$ for any odd $n$.

Whenever $G$ is not abelian, use the tactic provided by Lord Shark the Unknown: the inner automorphisms of $G$ are a subgroup of $\mathbb{Z}_n$ and are hence cyclic. But $\operatorname{Inn}G \cong G/Z(G)$, so $G/Z(G)$ is cyclic, which implies that $G$ is abelian, a contradiction.

Robin Aldabanx
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  • This map $\alpha$ is trivial if every element has order two. Showing that any abelian group of order $\geq3$ has an automorphism of order $2$ is non-trivial and relies on choice (unless $G$ is assumed to be finitely generated, which it is not here). See here for the appropriate working. – user1729 Nov 20 '19 at 09:24