Originally Posted by
Drexel28 I just helped someone with this in another thread.
Just note that if $\displaystyle a,b\in G'$ then $\displaystyle a=\theta(g),b=\theta(g')$ and so $\displaystyle ab=\theta(g)\theta(g')=\theta(gg')=\theta(g'g)=\th eta(g')\theta(g)=ba$