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.
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.
take $\displaystyle a,b\in G$ and $\displaystyle a\not =b$. Then $\displaystyle f(a)f(b)=f(ab)=(ab)^{-1}=b^{-1}a^{-1}=f(b)f(a)$. But f is an isomorphism (in particular an automorphism), so it has an inverse which is also an isomorphism from G to G.
$\displaystyle f^{-1}(f(a)f(b))=f^{-1}(f(b)f(a))$
$\displaystyle f^{-1}(f(a))f^{-1}(f(b))=f^{-1}(f(b))f^{-1}(f(a))$
$\displaystyle ab=ba$
Given $\displaystyle g,h \in G, \; (gh)^{-1} = h^{-1} g^{-1} $ since $\displaystyle (gh)(h^{-1} g^{-1}) = (h^{-1} g^{-1})(gh) = 1 $.
But we know $\displaystyle f $ is an isomorphism and hence a homomorphism, so $\displaystyle f(gh) = f(g)f(h) $.
Hence $\displaystyle h^{-1} g^{-1} = (gh)^{-1} = f(gh) = f(g)f(h) = g^{-1} h^{-1} $.
$\displaystyle \exists \; a,b \in G $ s.t. $\displaystyle a = g^{-1}, \; b = h^{-1} $.
Therefore $\displaystyle ab = ba \; \forall a,b \in G $ which means $\displaystyle G $ is abelian.