Finding a contradiction to $G$ being abelian when $G\cong M\times L$

Question

I am trying to disprove/prove the following:

"Let $M, L \subseteq G$, $G \cong M \times L$. Then if all the elements of $G$ commute with all the elements of $L$, we have that $G$ is abelian."

For this to be true, it would be necessary that $(ab, cd) = (ba, dc)$,$\ \ \forall a,b \in M, \forall c,d \in L$. As we don't necessarily have $ab=ba$ (since we don't know if $M$ is abelian, then I believe this statement would be false). However I cannot find a counterexample to show this so I'm not sure if I'm perhaps missing out on something.

So I'm just wondering if there's a neat way to disprove/prove the statement, since I can't really seem to find a counter example.

Answer 1

Let $L=\{e\}$ and $M=D_3$, the dihedral group of order $6$.