# Abstract Algebra: define an operation

murmillo

## Homework Statement

Does the rule g*x = xg^-1 define an operation of G on G?

## The Attempt at a Solution

I don't even know what this means. Could someone just tell me what it means for a rule to define an operation of one group on itself? I should be able to figure it out from there.

## Answers and Replies

spamiam
murmillo
Oh, I see. OK, thanks, I can take it from here.

Homework Helper
Gold Member
So you have to prove that (gh)*x=g*(h*x)
$$(gh)*x=x(gh)^{-1}=x(h^{-1}g^{-1})=(xh^{-1})g^{-1}=(h*x)g^{-1}=g*(h*x)$$