Conjugation question

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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.

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