# Let $N, K$ be abelian and subgroups of $G$, with $N$ normal and $G=NK$. Is $G\cong N\times K$?

Problem at hand (not homework): Let $N, K$ be subgroups of a group $G$, with $N$ normal in $G$. If $N$ and $K$ are abelian groups and $G= NK$, is $G$ the direct product of $N$ and $K$?

So, here's the criteria that I know:

1. Given normal subgroups $N_1, \cdots N_k$ of $G$ such that any element $g\in G$ can be written uniquely as $g=n_1\cdots n_k$, where $n_i\in N_i$. Then, $G\cong N_1\times\cdots\times N_k$.

2. If $M, N\trianglelefteq G$ such that $G=MN$ and $M\cap N=\{e\}$, then $G=M\times N$.

I think that conjecture is false, but I don't know how to go about utilizing the facts that I've listed above. I looked at the proofs of the above theorems (we used (1) to prove (2)), and they use normality.

$S_3$ has a subgroup $K$ of order $2$ and a subgroup $N$ of order $3$, and you can verify that the product $HK = S_3$ (their intersection is $\{e\}$ and the product is a subgroup with at least $5$ elements). Moreover, because $N$ has index $2$ in $S_3$, it is automatically normal. However, $S_3\not\cong H\times K$, as $S_3$ is not abelian.
This is false. Let $G=A_4$ and let $N$ be the Klein-four group inside $A_4$ represented as $\{e,(12)(34),(13)(24),(14)(23)\}$ so that $N\trianglelefteq G$ and let $K=\langle (123)\rangle$. We know that $|NK|=\frac{|N||K|}{|N\cap K|}=12=|A_4|$ so that $NK=A_4$ yet the direct product of $N$ and $K$ must be abelian as each is individually abelian, yet $A_4$ is not abelian.