The question is the same as here, and I understand the proof in it.
But I don't know why do we need both $H, K\mathrel{\unlhd}G$? I think just one of them being a normal subgroup will be enough.
W.L.O.G let $H\mathrel{\unlhd}G$ Then $hk=(khk^{-1})k=kh$, and the rest are the same as the post.
Where did I go wrong?
Thank you in advance for any help.
The equality $hk = khk^{-1}k$ seems wrong, unless $k$ and $h$ commute - but this is exactly what needs to be shown.
You need both $H$ and $K$ to be normal in $G$ because in a direct product every element of $H$ commutes with every element of $K$. You can show that this is equivalent to asking that both $H$ and $K$ are normal in $G$. If you only ask that $H$ is normal, you get what is called a semidirect product of $H$ and $K$, denoted $H \rtimes K$. To see this in action. Consider the cyclic group $\mathbb{Z}_{6}$ of order six. You can show that $\mathbb{Z}_{6} \cong \mathbb{Z}_{3} \times \mathbb{Z}_{2}$. On the other hand, the symmetric group on three letters is $S_{3} \cong \mathbb{Z}_{3} \rtimes \mathbb{Z}_{2}$, and you can check here that $\mathbb{Z}_{2}$ is not normal in $S_3$.
Here are Wikipedia links to this:
https://en.wikipedia.org/wiki/Direct_product_of_groups and https://en.wikipedia.org/wiki/Semidirect_product