Suppose that $G$ is a commutative group, and define $f∶G → G$ by $f(a) = a^{−1}$ . Prove or disprove: $f$ is an isomorphism.

Question

I think it is isomorphic but I don't how to actually prove it. I think it is because $a$ would go to $a^{-1}$, $b$ to $b^{-1}$, $e$ to $e$ etc.

Answer 1

Let $f$ send $a$ to $a^{-1}$. Therefore, $$f(ab)=(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}=f(a)f(b),$$ so that $f$ is an homomorphism (that's why we must consider $G$ as an abelian group). To prove that $f$ is injective, let $f(a)=f(b)$. Therefore, $a^{-1}=b^{-1}$. Because one element can't have more than one inverse, $a=b$. To prove it is surjective, let $a$ be any element of our group. Clearly, $f(a^{-1})=a$, so that the image of $f$ is the whole group. As a consequence, we have an isomorphism.

Answer 2

To prove whether or not a function is an isomorphism, you need to answer 2 questions : is it a homomorphism and is it bijective?

Homomorphism :

Prove that for any $a,b \in G$ you have $f(a\cdot b) = f(a)\cdot f(b)$

In order to prove that, you will need the fact that G is commutative.

Bijection :

There are many ways to show this. You could prove that $f$ is invertible OR that $f$ is injective and surjective.