Suppose G is a finite group, H and K normal subgroups, gcd(|H|, |K|) = 1, and |G| = |H| |K|. Prove $G \cong H \times K$.
I have a proposition that say if H and K are normal subgroups of G, $H \cap K = \{e\}$, and G = HK, then $G \cong H \times K$. My question is if this is the proposition that I need to use and how to go about showing that $H \cap K = \{e\}$, and G = HK, or how else can I go about proving this.
As in the David's comment , you will get that since $|H \cap K| | H;K$ and $gcd(|H|;|K|)=1$, their intersection is the identity elementy. Since the intersection of two normal subgroups $H$ and $K$ is $\{aba^{-1}b^{-1}| a \in H; b \in K\}$ we get that the elements of $H$ and $K$ commute and using the fact that $|G|=|H||K|=\frac{|H||K|}{1}=\frac{|H||K|}{|H \cap K|}=|HK|$, we have that $G=HK$ and hence $G= H \rtimes K $ and from the commutativity of the elements of $H$ and $K$, we have that $G \cong H \times K$.
For the first part see @David C. Ullrich's comment...
For the second part, of course $HK\subset G$, and then $\mid HK\mid=\mid G\mid$, since $\mid HK\mid=\frac{\mid H\mid \mid K\mid}{\mid H\cap K\mid}$. See this... So $HK=G$.