# Show that $Aut(G) \leq S_{G}$, the symmetric group.

Could someone let me know if I did anything wrong in my answer?

Show that $Aut(G) \leq S_{G}$, the symmetric group.

Subset: Since $Aut(G)$ consists of functions that map elements from $G$ to $G$, these functions, which are bijective due to $Aut(G)$ being a set of isomorphisms, are permutations of the set $G$. Therefore, $Aut(G)\subset S_{G}$.

Identity: Both $Aut(G)$ and $S_{G}$ have an identity function that maps an element to itself. Since $Aut(G)\subset S_{G}$, it must be that that identity function is in $Aut(G)$.

Inverse: Since $Aut(G)$ is a set of isomorphisms, by definition, an inverse exists for each isomorphism.

Closure: Any bijective function composition with domain $G$ and codomain $G$ will be a bijective function with domain $G$ and codomain $G$. The operation is automatically preserved due to it being an isomorphism, and so the operation is closed.

• Looks good. If you know the characterisation of subgroups this can be shortened, but the ideas that are used in this characterisation are used in your proof and making things explicit is never a bad idea. – user370967 Mar 5 '18 at 17:05
• For the inverse, you actually need to show that the inverse is a morphism. For the closure, the operation is automatically preserved due to it being a morphism. – Clément Guérin Mar 6 '18 at 1:10