If $\lvert H\oplus K \rvert = \lvert G\rvert$, then $G\simeq H\oplus K$? Background: I have a follow up question to my question here: Groups of order $pq$ are cyclic
Ultimately I am wanting to prove that a group of order $pq$ where $p\nmid q-1$ is cyclic. I have reduced the case to needing to show that the group is Abelian. The key is that I don't know the Sylow Theorems.
So now a related question that I would like to see if is true is:
Questions: Let $G$ be a finite group. Let $H$ and $K$ be two subgroups such that $H\cap K = \{e\}$ and $HK = G$ (or $\lvert H\rvert\cdot \lvert K \rvert = \lvert G\rvert$). I am wondering if it is true that $G \simeq H\oplus K$ (external direct product).
I am thinking this is true when I think about examples like $A_n \leq S_n$ and $\{(12)\}\leq S_n$. 
 A: No, because $S_3\not\cong C_2\oplus C_3$.
A: 
Questions: Let $G$ be a finite group. Let $H$ and $K$ be two subgroups such that $H\cap K = \{e\}$ and $HK = G$ (or $\lvert H\rvert\cdot \lvert K \rvert = \lvert G\rvert$). I am wondering if it is true that $G \simeq H\oplus K$ (external direct product).

Nope. You must have that both $H,K$ are normal in $G$ to get the result.
So it would be like this.

Let $G$ be a finite group. Let $H$ and $K$ be two normal subgroups of $G$ such that $H\cap K = \{e\}$ and $HK = G$ (or $\lvert H\rvert\cdot \lvert K \rvert = \lvert G\rvert$). Then $G \simeq H\oplus K$ (external direct product).

For example, $\mathbb{Z}_6\simeq\mathbb{Z}_2\oplus\mathbb{Z}_3$.
More in general you can say something like this:

Let $G$ be a finite group. Let $H$ and $K$ be two subgroups of $G$ such that $H\triangleleft G$, $H\cap K = \{e\}$ and $HK = G$ (or $\lvert H\rvert\cdot \lvert K \rvert = \lvert G\rvert$). Then $G \simeq H\rtimes K$ (external semidirect product).

For example $D_3\simeq \mathbb{Z}_3\rtimes \mathbb{Z}_2$.
Note the order when writing the semidirect product, written like that the normal subgroup must go to the left.
