1. ## normal subgroup

Assume H and K are normal subgroups of G with the intersection of H and K is equal to 1. Prove xy = yx for every x in H and y in K. (That is, to show $x^{-1}y^{-1}xy$ is in the intersection of H and K)

2. Originally Posted by dori1123
Assume H and K are normal subgroups of G with the intersection of H and K is equal to 1. Prove xy = yx for every x in H and y in K. (That is, to show $x^{-1}y^{-1}xy$ is in the intersection of H and K)
$x(yx^{-1}y^{-1}) \in H \text{ and }(xyx^{-1})y^{-1} \in K$.
Therefore, $xyx^{-1}y^{-1} = e$ and so $xy = yx$.

3. Originally Posted by ThePerfectHacker
$x(yx^{-1}y^{-1}) \in H \text{ and }(xyx^{-1})y^{-1} \in K$.
How do you know $x(yx^{-1}y^{-1}) \in H$?

4. Originally Posted by dori1123
How do you know $x(yx^{-1}y^{-1}) \in H$?
Because $x \in H$.
Because $yxy^{-1} \in H$ -- this is by normality.
Therefore $x(yxy^{-1}) \in H$ by closure on $H$.