# Comparing definitions of internal direct product

The internal direct product is usually defined by either set of the requisites shown below.

Let $G$ be a group and $H,K\leq G$ satisfying the following conditions:

• $G=HK=\left\{ hk\mid h\in H,k\in K\right\}$

• $H\cap K=\{e\}$

• $hk=kh$, $\forall k\in K$ and $\forall h\in H$.

or

• $G=\left\langle H\cup K\right\rangle$

• $H\cap K=\{e\}$

• $H,K\trianglelefteq G$

Then $G$ is the internal direct product of $H$ and $K$.

My goal is to compare both definition. Firstly I wanted to prove the following:

If $H\cap K=\{e\}$ then $H,K\trianglelefteq G\iff hk=kh,\forall k\in K\,\forall h\in H$

($\implies$) One must prove that $[h,k]\equiv hkh^{-1}k^{-1}=e$. On the one hand, by definition of $K\trianglelefteq G$ one has $\underset{\in K}{\underbrace{hkh^{-1}}}k^{-1}\in K$. On the other hand, by definition of $H\trianglelefteq G$, one has $\underset{\in H}{h\underbrace{kh^{-1}k^{-1}}}\in H$. Since $[h,k]\in H\cap K$ it follows that $[h,k]=e$.

($\impliedby$) Conversely I couldn't give a proof or a give a counter example. I think the idea of a proof I had fails by the following reasoning: By $[h,k]=e$ one gets that $hkh^{-1}=k$. For $g\in H$ it is the case that $gkg^{-1}=k\in K$ and for $g\in K$ one gets $gkg^{-1}\in K$. However for the case that $g\in (G\setminus H )\cap( G\setminus K)$ I can't verify the conjugation closure for $K$. In other words, $H\cup K$ isn't a group in general unless one of them is contained in the other (here it isn't the case) so clearly $G\neq H\cup K$.

Question 1: Is there a way to prove $\impliedby$? If not, could you give me a hint to provide a counter example?

Now I want to check that the first item in both definitions are the same.

$\left\langle H\cup K\right\rangle =\left\{ \text{"all words in }H\cup K\text{"}\right\} =\left\{ {\prod}x_{i}^{\pm1}\mid x_{i}\in H\,\lor\,x_{i}\in K\right\}$. Now by what I saw above the elements in $H$ and $K$ commute thus $$\left\langle H\cup K\right\rangle =\left\{ {\prod}h_{i}^{\pm1}{\prod}k_{i}^{\pm1}\mid h_{i}\in H,k_{i}\in K\right\} =\left\{ hk\mid h\in H,k\in K\right\} =HK$$

Question 2: Is the above proof correct?

For the first part you have to include the fact that every element of $G$ can be written as a product of elements of $H,K$, otherwise the second direction isn't true.
If $G \not = \langle H,K \rangle$ then the statement isn't true. For a counterexample consider the group $G = S_3 \times S_3$. Now take the subgroup $H =\langle ((12),e) \rangle$, $K = \langle (e,(13)) \rangle$. We have that their intersection is trivial and obviously we have that the elements commute by the definition of direct product of groups. On the other side both groups aren't normal in $G$, as after all they aren't normal in the respective copies of $S_3$.
The second proof is correct. After all it's trivial that $HK = \langle H,K \rangle = \langle H \cup K \rangle$, as $N,K \unlhd G$