Proving a group $S$ is isomorphic to a single permutation group of $S$

Question

I am learning introductory group theory and want to prove the following:

Let $S:(S,*)$ be a group and $G:(G,*)$ be a group of whose elements are a result of a bijection from $S\to S$.

Prove $S\cong G$

Work: $G$ and $S$ are isomorphic if through a bijection $\gamma$, $\gamma(a*b)=\gamma(a)*\gamma(b)$ for $\forall a,b\in S$ and $\forall \gamma(a),\gamma(b)\in G$.

Let $a*b=c$. By the definition of groups, $c\in S$. Since $G$ contains the same elements as $S$, $\gamma(a)*\gamma(b)\in G,S$

Thus we can let $\gamma(a)*\gamma(b)=\gamma(c)$, which completes the proof.

However, this seems wrong so if anyone can help me that would be greatly appreciated.

Answer 1

I'm guessing this is what you actually mean:

Let $\gamma_a(s) := as \ \forall s \in S$. Then $(\{\gamma_a : a \in S\}, \circ) \simeq (S, *)$

Then of course $f: S \to G, f(s) = \gamma_s$ is an isomorphism.