A group $G$ whose automorphism group contains a subgroup isomorphic to $G$

Question

The problem is to find a non-trivial group $G$ whose automorphism group contains a subgroup isomorphic to $G$.

I know that $\operatorname {Aut} S_3\simeq S_3$, so the first idea is to prove this. $S_3$ acts on itself by left multiplication, and this action is faithful. This gives an injective group homomorphism $S_3\to \operatorname {Perm} S_3$ from $S_3$ to the group of permutations on the set $S_3$. At this point, I have two questions:

• Is $\operatorname {Perm} S_3$ the same as $\operatorname {Aut} S_3$ (if so, in what sense)? $\operatorname {Aut} S_3$ is the group of isomorphisms from $S_3$ to $S_3$, and it's not evident for me that they are "the same".
• I guess it suffices to show (apart from the above) that $\operatorname {Perm} S_3$ has order $6$. How do I do this?

Answer 1

Answer to the question in your post: take $G$ with trivial center, then $G/Z(G)=G \cong Inn(G)$ is a subgroup of $Aut(G)$. So $S_3$ is a good example.

Answer 2

No. A permutation of $S_3$ (i.e., of its underlying set) is not an automorphism of $S_3$. For example, every automorphism of a group leaves the neutral element fixed - and left multiplication won't (except in the trivial case).

However, note that $S_3$ has exactly three elements of order $2$ and any automorphism of the group $S_3$ must map elements of order $2$ to elements of order $2$. Can you exploit this fact (and that $S_3$ is generated by its elements of order $2$) to see how $\operatorname{Aut}S_3\cong S_3$?

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By the way, with $G=1$, youÄd also have $\operatorname{Aut} G\cong G$.