I'm new to group theory and have some struggles to show the the question. Here, a subgroup $H\subset S_n$ is called transitive if for every $\{i,j\}\subset\{1,2,\dots,n\}$ there is a $\tau\in H$ with $\tau(i) = j$ and Cayley's theorem says that every group $G$ is isomorphic to a subgroup of $S_G$.
We can define a homomorphism $\varphi : G\to S_G$ and know the fact that $G$ is isomorphic to $\varphi(G)$. So I think that we need to show that $\varphi(G)$ is a transitive subgroup of $S_G$ (if I'm correct so far) and I cannot progress anymore. Can you help me to show this or if my approach is wrong, correct it please.
You have to chase the proof of Cayley's theorem. If your group $G$ is $$ \{e=g_1, g_2, \ldots, g_n\} $$ then we have the permutation $\sigma_k$ induced by $g_k$ which respects the permutation on the subscripts obtained by left-multipying by $g_k$. So, if $g_kg_i = g_j$ then $\sigma_k(i)=j$. Then $g_k \mapsto \sigma_k$ is an isomorphism from $G$ to the subgroup $\{\sigma_k \mid k = 1, 2, \ldots, n\}\subseteq S_n$.
Now the result is clear: if you want to know which permutation maps $i$ to $j$, you just use the subscript $k$ corresponding to $g_jg_i^{-1}$. This is some element $g_k \in G$, and by definition $\sigma_k(i)=j$.
Edit: I see that I did this in $S_n$ and not $S_G$, but it's the same idea: the equation $xi=j$ can always be solved for $x \in G$.