3
$\begingroup$

Let $G$ be any group, and let $\phi:G\to G$ be given by $\phi(g)=g^{-1}$. Is $\phi$ a homomorphism?

My answer thus far:

I just start by seeing if the homomorphism property $\phi(ab)=\phi(a)\phi(b)$ is satisfied:

$\phi(g_1g_2)=(g_1g_2)^{-1}=g_2^{-1}g_1^{-1}=\phi(g_2g_1)=\phi(g_2)\phi(g_1)$

(EDIT: Added the last part, $=\phi(g_2)\phi(g_1)$)

So... $\phi$ is a homomorphism, but only if G is an abelian group? I'm probably doing something wrong here.

$\endgroup$
1
$\begingroup$

In general, $\phi$ is not a homomorphism. We only need to consider a non-abelian group with $a,b$ such that $ab\neq ba$.

Then if $\phi$ is a homomorphism we have $\phi(ab)=b^{-1}a^{-1}=a^{-1}b^{-1}$

or $ab=ba$, a contradiction.

$\endgroup$
6
$\begingroup$

You are exactly right, except you should have written: $$\cdots = g_2^{-1}g_1^{-1} = \phi(g_2)\phi(g_1)$$ In general $\phi(g) = g^{-1}$ defines an isomorphism from $G$ to its opposite group $G^\text{op}$, which has the same underlying set as $G$, but his operation is ``reversed''. Ie, if the operation of $G$ is $\cdot$ and the operation of $G^{\text{op}}$ is $*$, then $b*a := a\cdot b$.

Indeed as you observed $\phi$ is only an automorphism if $G$ is abelian, which is precisely when $G^\text{op} = G$.

$\endgroup$
  • $\begingroup$ Oop, dang I thought I did write that. I guess I stopped short. Thanks! $\endgroup$ – Indigo Nov 5 '15 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.