How is injectivity enough to show function is an isomorphism

Question

I have some problem understanding one part of the proof of Cayley's theorem.

The theorem states that if $G$ is a group, then $G$ is isomorphic to a group of permutations. The first part of the proof is to use an earlier theorem stating that there exists a homomorphism $\phi :G\rightarrow S_G$ where $(\phi(a))(x)=ax$, $x\in G$. Then they show that $ker(\phi)=\{e\}$, which implies that the functions is injective. This is the end of the proof.

How does this show that $\phi$ is an isomorphism? Don't we need to show that it is surjective as well, or is there something I am missing?

Answer 1

Once you show that $\phi$ is injective, there are two more steps which are hidden:

1. Show (or recall) that $\phi(G)$ is a subgroup of $S_G$.
2. Show that $\phi: G\to\phi(G)$ is an isomorphism.

Note that $\phi: G\to S_G$ is not an isomorphism, but $\phi: G\to\phi(G)$ is an isomorphism. While this may seem strange at first, remember that a function is defined not only by its action, but also by its codomain and domain. The two functions above differ in their codomains, and have different properties because of that.