suppose $G$ and $H$ are groups and let $f: G\rightarrow H$ with $f(g)=h$ be a group homomorphim. Is it true that $f(g^{-1})=h^{-1}$?

Question

Recently I have began to learn about homomorphims in my introductory abstract algebra course and I had been working on a homework assignment that, if this statement were true, would make it much easier solve. I am certain that this is true since if we assume that $f(g^{-1})=h'\neq h^{-1}$ then $f(gg^{-1})=f(e_{G})=e_{H}$, but also that $f(gg^{-1})=f(g)f(g^{-1})=hh'\neq e_{H}$ which is a contradiction. Thus it must be the case that if an $x$ from the domain maps to a $y$ in the range then $x^{-1}$ must map to $y^{-1}$. Please let me know if there is anything wrong with what I have said here or if this statement is even true at all.

Answer 1

Your argument is good, but a bit contorted: no contradiction is needed.

If in a group we have $x=x^2$, then $x=e$, because we have $xx^{-1}=x^2x^{-1}$ and so $e=x$.

Consider now $f(e_G)=f(e_Ge_G)=f(e_G)f(e_G)$. By the argument above, $f(e_G)=e_H$.

Next, from $$ e_H=f(e_G)=f(gg^{-1})=f(g)f(g^{-1}) $$ we conclude that $f(g^{-1})=(f(g))^{-1}$. The argument is easy: if $e=xy$, then $x^{-1}e=x^{-1}xy$ and therefore $x^{-1}=y$.

Answer 2

You can say this : $f(g)f(g^{-1})=hf(g^{-1})$. But $f(g)f(g^{-1})=e$. And now you can conclude.