# There does not exist group $G$ such that ${\rm Aut}(G)\cong \mathbb{Z}_n$ (for odd $n$)

I had this "almost bonus" question on the final in Group Theory recently: prove that there is no such group $$G$$ which would satisfy $${\rm Aut}(G)\cong \mathbb{Z}_n$$, where $$n$$ is an odd integer. I don't have much certainty if this proof is OK, and one's opinion would be appreciated.

Here's my attempt:

Suppose that there exists such a group $$G$$ that satisfies the above condition. Since $$\mathbb{Z}_n$$ is cyclic, $${\rm Aut}(G)$$ is also cyclic and $$G$$ is abelian, which implies that $${\rm Inn}(G)\cong \{e\}$$. Thus $${\rm Aut}(G)={\rm Out}(G)$$. (I don't think this fact is important here though). Now, since an automorphism sends a generator to a generator, and since each automorphism is completely determined by such mapping, $$|{\rm Aut}(G)|$$ must be of factorial order. But an integer factorial is always even. However, $$|\mathbb{Z}_n|$$ is odd, which is a contradiction. Therefore, no such $$G$$ exists.

I'm afraid that my proof is not very nice, but at least a genuine attempt was made.

• $n$ is assumed to be odd. Sorry, edited. Dec 13, 2015 at 5:25
• Note that we should have $n\neq 1$, otherwise the groups of order $1$ and $2$ are trivial counterexamples. Dec 13, 2015 at 20:32
• Aug 17, 2016 at 10:03

$\DeclareMathOperator{\Aut}{Aut}$ We'll show the slightly stronger result that $|\Aut(G)| \not= 2n+1$ for any $n \ge 1$ EDIT: Provided $G$ is abelian: see the comments.

If $\Aut(G)$ is trivial, then there is nothing to show. Suppose, then, that $G$ has a nontrivial automorphism group. We will show that $|\Aut(G)|$ is even by explicitly exhibiting an automorphism $\phi$ of $G$ having order $2$ for any group $G$. It will then follow by Lagrange's theorem that $\Aut(G)$ has even order.

If $G$ has an element of order at least $3$, then $\phi:x \to x^{-1}$ is the desired automorphism of even order. It only remains to consider the case where $G$ contains only elements of order less than or equal to $2$. In that case, we must have that $G$ is a vector space over $\mathbb{Z}_2$ (see Group where every element is order 2 for a proof of this result). If $G$ is one dimensional as a vector space, then $\Aut(G) = \Aut(\mathbb{Z}_2)$ is trivial, so there is nothing more to show. For $G$ having dimension greater than $1$, we have that $\Aut(G) \cong \operatorname{GL}(G)$, so if we pick an ordered basis for $G$, then the automorphism which interchanges the first and second basis vectors (and fixes all others) is an order $2$ automorphism. (This argument, which Slade gave in the comments, applies even to infinite $G$, while my previous argument tacitly assumed $G$ was finite.)

• This assumes that $G$ is finite, but it is very close to applying to infinite groups as well. An abelian group in which every nonzero element has order $2$ is a vector space over $\mathbb{Z}_2$, so it has a basis and we can write $G = \oplus_i \mathbb{Z}_2$ and find lots of nontrivial automorphisms. Dec 13, 2015 at 20:31
• @Slade Good point! I've changed my answer slightly to incorporate your comment.
– user88319
Dec 13, 2015 at 20:39
• @Strants : So you are assuming $G$ is abelian ?
– user228169
Apr 30, 2016 at 7:41
• Every group has an element of order at most 2 (namely, the identity element). You mean a group whose every element has order at most 2 (i.e. "exponent" 2). As a matter of fact, there do exist finite groups $G$ whose automorphism group ${\rm Aut}(G)$ has odd order. The correct proof of OP's desired claim begins as follows: if ${\rm Aut}(G)$ is cyclic then so is ${\rm Inn}(G)\cong G/Z(G)$, and a standard exercise shows that if $G/Z(G)$ is cyclic then $G$ is abelian.
– anon
Apr 30, 2016 at 19:14
• @arctictern Oops, you are correct. The requirement that $G$ is abelian is necessary if $G$ has an element of order greater than $2$: otherwise, $x \to x^{-1}$ will fail to be an automorphism. I've re-edited to make it clear I'm assuming that $G$ is commutative.
– user88319
Apr 30, 2016 at 19:20

It is not true that $|Aut(G)|=m!$ for some $m$. For example, if $G=\mathbb{Z}_5$, then $Aut(G)\cong\mathbb{Z}_4$ and 4 is not a factorial.

If you know that $G$ is abelian, then it is the product of cyclic group $G=\prod \mathbb{Z}_{p_i^{k_i}}$ where the $p_i$ are prime. It follows that $Aut(\mathbb{Z}_{p_i^{k_i}})$ is a subgroup of order $(p_i-1)p_i^{k_i-1}$ of $Aut(G)$. If $p_i$ is odd, then $p_i-1$ is even so the order of $Aut(G)$ is even. Similarly, if $p_i=2$ and $k\geq 2$ then $p^{k-1}$ is even. So you just need to check the case where $G= \mathbb{Z}_2^m$ in which unless $m=1$ (and then $Aut(G)$ is trivial), you can always find an automorphism of order 2 - for example, switch two basis elements.

Actually, it is true that $Aut(G)$ is cyclic (and not trivial) iff $G$ is cyclic of order 4, $p^k$ or $2p^k$ where $p$ is prime (see here). Using that it is easy to see that the order of $Aut(G)$ must be 2 or $(p-1)p^{k-1}$ both of which are even.

• This assumes that $G$ is finite, but the question appears to be valid even for infinite $G$. Dec 13, 2015 at 20:31

Suppose that $n\geq 3$ and $n$ is odd. If $Aut(G)\cong \mathbb{Z}_n$ then $Aut(G)$ is cyclic, which implies that $G$ is abelian. But if $G$ is abelian then the inversion map $x\mapsto x^{-1}$ is an automorphism of order $2$. By Lagrange's theorem, this implies that $|Aut(G)|$ is divisible by $2$. This is a contradiction since $n$ is odd.

• The inversion map may also have order one (but of course then there are other involutionary automorphisms).
– Dune
Dec 7, 2017 at 16:20
• For the klein's 4 group inversion mapping is same as identity map Nov 18, 2023 at 20:16