Prove that for each element $a$ in $G_{1}$, $f(a^{-1})=[f(a)]^{-1}$

Question

Prof. Pinter's "A Book of Abstract Algebra" presents this exercise:

Let $G_1$ and $G_2$ be groups, and let $f: G_1 \rightarrow G_2$ be an isomorphism.

Prove that for each element $a$ in $G_{1}$, $f(a^{-1})=[f(a)]^{-1}$. [Hint - you may use Theorem 2 of Chapter 4.]

This theorem states:

If $G$ is a group and $a$, $b$ are elements of $G$, then $ab=e$ implies $a=b^{-1}$ and $b=a^{-1}$.

Given that theorem, I'm not sure how to apply it to answer the exercise. Please give me a hint.

Answer 1

Assuming $f:G \to H$ is a homomorphism of groups, note that for any $a \in G$

$e_H = f(e_G) = f(aa^{-1}) = f(a)f(a^{-1}); \tag{1}$

now by the given theorem, renaming $a, b$ in its statement $f(a)$ and $f(a^{-1})$, respectively, we have

$f(a^{-1}) = [f(a)]^{-1} \tag{2}$

and so forth. QED!

Note: $f(e_G) = e_H$ since

$f(e_G) = f(e_G^2) = [f(e_G)]^2, \tag{3}$

whence

$e_H = [f(e_G)]^{-1} f(e_G) = [f(e_G)]^{-1}[f(e_G)]^2 = f(e_G). \tag{4}$

End of Note.

Answer 2

We have $e_2=f(e_1)=f(aa^{-1})=f(a)f(a^{-1})$, because $f$ is a group homomorphism. Now apply Theorem $2$.