# For finite group $G$ when is $|Aut(G)| < |G|$?

If $G$ is a finite group then we know $|Aut(G)|$ divides $(|G|-1)!$ ; I want to ask , if $G$ is a finite group with more than one element then is it true that $|Aut(G)| < |G|$ ( I know it is always true for cyclic groups) ? If $|Aut(G)| < |G|$ is not always true then can we characterize those groups for which it is true ?

• Well this is certainly true for all cyclic groups other than the trivial one because $\phi(n)<n$ for all $n\geq 1$. – RougeSegwayUser Oct 30 '14 at 4:28
• The automorphism group of the quaternion group is $S_4$, so it fails for this. – Peter Huxford Oct 30 '14 at 4:32
• Also, this is not the case for Alb. For this category, the problem amounts to enumerating the set of all $\mathbb{Z}$-bases. Even for $\mathbb{Z}_3^2$, this fails. It holds for $\mathbb{Z}_2 \times \mathbb{Z}_n$. In that case the number of $\mathbb{Z}$-bases is $\phi(n)$. Here I'm using Alb for the category of finite abelian groups (or $\mathbb{Z}$-modules). I'm fairly certain that this fails for all other finite abelian groups – RougeSegwayUser Oct 30 '14 at 4:42
• The automorphism group of $\Bbb{Z}_2\times\Bbb{Z}_2$ has order six. The automorphism group of $\Bbb{Z}_2^3$ has order 168... – Jyrki Lahtonen Oct 30 '14 at 5:02
• If the centre $Z(G) = 1$, then $G \cong \text{Inn}(G) \le \text{Aut}(G)$, so $|Aut(G)| \ge |G|$ in this case. e.g. all finite simple groups. – Alastair Litterick Oct 30 '14 at 20:17

This claim is false in general. You can check yourself that it fails for group of small order (quarterion group, for example). What is true is the order of elements in the automorphism group must be less than the order of the group itself. For a reference see here:

https://mathoverflow.net/questions/1075/order-of-an-automorphism-of-a-finite-group

For proof of the counter-example see here:

Automorphism group of the quaternion group