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 $\displaystyle x^{-1}y^{-1}xy$ is in the intersection of H and K)
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 $\displaystyle x^{-1}y^{-1}xy$ is in the intersection of H and K)
$\displaystyle x(yx^{-1}y^{-1}) \in H \text{ and }(xyx^{-1})y^{-1} \in K$.
Therefore, $\displaystyle xyx^{-1}y^{-1} = e$ and so $\displaystyle xy = yx$.
How do you know $\displaystyle x(yx^{-1}y^{-1}) \in H $?
Because $\displaystyle x \in H$.
Because $\displaystyle yxy^{-1} \in H$ -- this is by normality.
Therefore $\displaystyle x(yxy^{-1}) \in H$ by closure on $\displaystyle H$.