Actually I have solved the part "$\operatorname{Inn}(S_3)\simeq S_3$" by using the result-
"Let $G$ be a group. Then ${G \over Z(G)}\simeq \operatorname{Inn}(G)$, where $Z(G)=\{g\in G\mid gx=xg\: \forall x \in G\}$"

So I put $G=S_3$ and hence $Z(S_3)=\{e\}$, where $e$ is identity permutation. And thus ${S_3 \over Z(S_3)}\simeq S_3$
Finally, using the above result, we get, $\operatorname{Inn}(S_3)\simeq S_3$.
But how to prove $\operatorname{Aut}(S_3) \simeq S_3$. Although, One thing I have notice that, to prove $\operatorname{Aut}(S_3) \simeq S_3$, it is enough to show that $\operatorname{Inn}(S_3)\simeq \operatorname{Aut}(S_3)$. But I am stuck with it also. Can anyone suggest me a clear and rigorous wayout?
Thanks for your help in advance.

  • 2
    $\begingroup$ Let $\sigma\in{\rm Aut}(S_3)$, consider the image of say $(1,2)$ and $(1,3)$ (which together generate $S_3$) and show that $\sigma$ is an inner automomorphism $\endgroup$ May 8, 2018 at 7:31

2 Answers 2


Without appealing to inner automorphisms; an automorphism of $S_3$ permutes the set of transpositions $$\{(1\ 2),\ (1\ 3),\ (2\ 3)\},$$ and any permutation of the set of transpositions uniquely determines an automorphism of $S_3$.


Automorphisms send conjugacy classes to conjugacy classes and preserve the order of the elements. In $S_3$, distinct conjugacy classes correspond to elements of distinct orders. Therefore, every automorphism of $S_3$ sends each conjugacy class (cycle type) to itself, so it is inner.


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