Finding groups $G$ such that $G \cong \mathscr{S}(G)$

Solution 1:

Interesting question! It seems that there is such a group. I have no idea about a classification of these groups, myself.

Solution 2:

I can help a little, maybe not if you already thought of it : there is no point in trying to find counter examples with finite groups ; for if $G \cong \mathscr S(G)$, since $\mathscr S(G) \cong G/Z(G)$, we have $G \cong G/Z(G)$, which means those two are in bijection. Since this means $|G| = |G|/|Z(G)|$, we have $|Z(G)| = 1$ and $Z(G) = \{ e \}$.

Now while looking at infinite groups it may be interesting that you are basically trying to either find a group $G$ with a non-trivial center such that $G \cong G/Z(G)$, or trying to prove that every group with a non-trivial center is such that $G \ncong G/Z(G)$. My first approach would be to try some nasty groups as counter-examples rather than trying proofs of the second choice...