What is known about automorphism group in general and about $|\text{Aut}(G)|$? Is it true that $|\text{Aut}(G)| \le |G|$? Exist any algorithm to build $\text{Aut}(G)$ for given $G$?

$G$ is finite.

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    $\begingroup$ $|\mathrm{Aut}(G)| \leq |G|$ is surely false in general considering that the automorphism groups of non-abelian simple groups contain a copy of said simple group (embeddig via conjugation action) and e.g. for the groups of type $A_n$ these embeddings are not surjective. $\endgroup$ Commented Sep 25, 2014 at 13:45

2 Answers 2


It is not true in general that $|{\rm Aut}(G)| \le |G|$. An elementary abelian group $C_p^n$ of order $p^n$ has automorphism group ${\rm GL}_n(p)$, which has order $$(p^n-1)(p^n-p)(p^n-p^2)\cdots (p^n-p^{n-1}).$$

There is an obvious upper bound $|{\rm Aut}(G)| \le |G|^{\log_2(|G|)}$, which could be improved slightly. You get the largest possible order of ${\rm Aut}(G)$ as a function of $|G|$ with $G=C_2^n$.

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    $\begingroup$ In case the obvious upper bound isn't obvious to the OP, I suggest that (s)he prove as an exercise that every finite group $G$ can be generated by $\log_2|G|$ of its elements. $\endgroup$ Commented Sep 25, 2014 at 13:49
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    $\begingroup$ $G=Z_2\times Z_2$ is good counter example, I think as $Aut(G)\cong S_3$ $\endgroup$
    – mesel
    Commented Sep 25, 2014 at 13:59

An automorphism of the symmetric group $S_6$ either maps the set of transpositions to itself (in which case the automorphism is inner) or maps each transposition to a product of three transpositions. It can be shown that $Inn(S_6) \cong S_6$ and $Inn(S_6)$ has index 2 in $Aut(S_6)$. Thus $Aut(S_6)$ has twice as many elements as $S_6$.


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