# Proof of a linear operator acting on an inverse of a group element

1. Oct 7, 2013

Hey guys!

Basically, I was wondering how to prove the following statement. Ive seen it in the Hamermesh textbook without proof, so I wanted to know how you go about doing it.

Let's say you have a group element $g_{1}$, which has a corresponding inverse $g_{1}^{-1}$. Let's also define a linear transformation D for this group element.

So what im trying to prove is that

$D(g_{1}^{-1}) = [D(g_{1})]^{-1}$

Can u guys point me in the right direction?

Thanks!

2. Oct 7, 2013

### Office_Shredder

Staff Emeritus
By linear transformation do you mean group homomorphism? If so just consider
$$D(g_1 g_1^{-1}) = D(g_1) D(g_1^{-1})$$

3. Oct 7, 2013

yes it is a homomorphism. But what u wrote in the previous message, where do I go from there? cos $D(g_{1}g_{1}^{-1})$ is just $D(E)$ where E is the identity, right? o.o

4. Oct 7, 2013

### Office_Shredder

Staff Emeritus
Yes, and what is D(E) equal to?

5. Oct 7, 2013

OHH I get it. Basically, because $D(E) = E$, we can say that
$E = D(g_{1})D(g_{1}^{-1})$
Then by multiplying both sides on the left by $[D(g_{1})]^{-1}$ we get
$[D(g_{1})]^{-1} = D(g_{1}^{-1})$....I think...