# Thread: Proof on abelian and isomorphic groups

1. ## Proof on abelian and isomorphic groups

If a function f, defined f(g)=g^-1 maps a group G to G, is isomorphic, then G is abelian.

How do I go about proving this one?

Edit: Sorry, wrong section, delete please.

2. take $a,b\in G$ and $a\not =b$. Then $f(a)f(b)=f(ab)=(ab)^{-1}=b^{-1}a^{-1}=f(b)f(a)$. But f is an isomorphism (in particular an automorphism), so it has an inverse which is also an isomorphism from G to G.

$f^{-1}(f(a)f(b))=f^{-1}(f(b)f(a))$

$f^{-1}(f(a))f^{-1}(f(b))=f^{-1}(f(b))f^{-1}(f(a))$

$ab=ba$

3. Given $g,h \in G, \; (gh)^{-1} = h^{-1} g^{-1}$ since $(gh)(h^{-1} g^{-1}) = (h^{-1} g^{-1})(gh) = 1$.

But we know $f$ is an isomorphism and hence a homomorphism, so $f(gh) = f(g)f(h)$.

Hence $h^{-1} g^{-1} = (gh)^{-1} = f(gh) = f(g)f(h) = g^{-1} h^{-1}$.

$\exists \; a,b \in G$ s.t. $a = g^{-1}, \; b = h^{-1}$.

Therefore $ab = ba \; \forall a,b \in G$ which means $G$ is abelian.