Show that If $\phi$ is a homomorphism, then $\phi(a^{-1}) = [\phi (a)]^{-1}$

Question

It makes sense that this is true, but I am having trouble coming up with a way to show that it is true. Any ideas? Thanks!

Answer 1

Your question is not very clear since the domain and codomain of your function $\phi$ is not defined.

Let $G$, $H$ be groups, $\phi:G\rightarrow H$ be a homomorphism.

$$\phi(1_G)=1_H$$

This can be done by considering the equation $\phi(1_G)=\phi(1_G1_G)=\phi(1_G)\phi(1_G)$.

For $a\in G$, $$\phi(a^{-1})=\phi(a)^{-1}$$

Since $\phi(a)\phi(a^{-1})=\phi(aa^{-1})=\phi(1_G)=1_{H}$ and also $\phi(a^{-1})\phi(a)=\phi(1_G)$, by definition of inverse of a group, your problem is solved.

Answer 2

Hint: $\phi(aa^{-1})=\phi(1)=1=\phi(a)\phi(a^{-1})$