Prove $f$ Matches Group $G_{1}$'s Neutral Element in Group $G_{2}$

Question

Dr. Pinter's "A Book of Abstract Algebra" poses the exercise:

Let $G_{1}$ and $G_{2}$ be groups, and let $f: G_{1} \rightarrow G_{2}$ be an isomorphism.

If $e_{1}$ denotes the neutral element of $G_{1}$ and $e_{2}$ denotes the neutral element of $G_{2}$, prove that $f(e_{1})=e_{2}$. [Hint - In any group, there is exactly one neutral element; show that $f(e_{1})$ is the neutral element of $G_{2}$.]

I'm stuck as to how to answer this question. Can someone please give me a hint?

Answer 1

Hint: For every $g\in G_1$,

$$f(g) = f(ge_1) = f(g)f(e_1).$$

Then use the fact that it is bijective to conclude.