Show that $|\phi(a)| \mid|G|$

By Lagrange, the order of $a$ divides the order of $G$. By the homomorphism property, the order of $\phi(a)$ divides the order of $a$. Now by transitivity of divisibility, the result follows.

You don't need injectivity.

However, injectivity could be considered to simplify the proof. Because $\phi(a)$ will be an element of the image. Then apply the first isomorphism theorem and Lagrange.

Let $\phi(G) = \{ \phi(g) : g \in G\}$ be the image of $G$ under $\phi$. It is a subgroup of $H$ (check this), and in particular, $\phi(G)$ is a group. Because $\phi(G)$ is a group containing the element $\phi(g)$, Lagrange's theorem tells us that the order of $\phi(g)$ divides the order of $\phi(G)$.

Now because $\phi$ is assumed to be one-to-one, what does that tell you about the relationship between the order of $G$ and the order of $\phi(G)$?