# Conjugation question

• April 21st 2010, 03:18 AM
Giraffro
Conjugation question
If $G$ is a group, $K \lhd G$ is proper and $\forall x \in G \backslash K, x^2 = e$, show that $\forall t \in G \backslash K, \forall a \in K, t^{-1} at = a^{-1}$.

I'm completely confused on how to solve this. I just seem to be going round in circles. (Headbang)

Can anyone help?
• April 21st 2010, 03:28 AM
Opalg
Quote:

Originally Posted by Giraffro
If $G$ is a group, $K \lhd G$ is proper and $\forall x \in G \backslash K, x^2 = e$, show that $\forall t \in G \backslash K, \forall a \in K, t^{-1} at = a^{-1}$.

If $a\in K$ and $t \in G \setminus K$ then $at \in G \setminus K$. Thus $(at)^{-1} = at$, and also $t^{-1} = t$.
• April 21st 2010, 03:36 AM
Giraffro
Thanks! I thought it was something blindingly obvious.