Originally Posted by

**thehobbit** I thought about this one while shopping and now the answer seems quite simple.

For (a), since $\displaystyle HK \leq G$ and $\displaystyle hk \in HK$, we have $\displaystyle |HK| = \frac{|H| \cdot |K|}{|H \cap K|}$, thus $\displaystyle (hk)^{|H| \cdot |K|} = (hk)^{|HK| \cdot |H \cap K|} = e^{|H \cap K|} = e$

For (b), picking up where you left off, $\displaystyle o(n) = o(k^{-1}h)\:\big|\:|H| \cdot |K|$. Since $\displaystyle o(n)$ is coprime to $\displaystyle |H| \cdot |K|$, we must have $\displaystyle n=1$, hence $\displaystyle N$ is the trivial subgroup, yielding the desired result.