Product group isomorphism

Question

Let $H$ and $K$ be subgroups of a group $G$, and let $f:H\times K \to G$ be the multiplication map, defined by $f(h,k)=hk$. Show that $f$ is an isomorphism from the product group $H \times K$ to $G$ if and only if $H \cap K=\{1\}$, $HK=G$, and also $H$ and $K$ are normal.

$\textbf{Proof:}$

"If" part:

Since $HK=G$, $f$ is surjective.

To show that $f$ is injective, notice that if $H \cap K=\{1\}$, then $$f(k_1,h_1)=f(k_2,h_2)\implies k_1h_1=k_2h_2$$ $$\implies k_2^{-1}k_1h_1h_1^{-1}=k_2^{-1}k_2h_2h_1^{-1}$$ $$\implies k_2^{-1}k_1=h_2h_1^{-1}$$ $$\implies k_2^{-1}k_1=h_2h_1^{-1}=1$$ $$\implies k_1=k_2, h_1=h_2$$

To show that $f$ is a homomorphism, notice that $$f((k_1,h_1)\cdot(k_2,h_2))=f(k_1h_1,k_2h_2)=k_1h_1k_2h_2$$

Since $K$ is normal, we have $$h_1k_2=k'_2h_1$$ Since $H$ is also normal, we have $$k'_2h_1=h'_1k'_2$$ Hence $$h_1k_2=h'_1k'_2$$ and therefore $${h'}_1^{-1}h_1=k'_2k_2^{-1}$$ Because $H\cap K=\{1\}$, we have $${h'}_1^{-1}h_1=k'_2k_2^{-1}=1$$ and hence $$h'_1=h_1,k'_2=k_2$$ Therefore, $$f((k_1,h_1)\cdot(k_2,h_2))=k_1k_2h_1h_2=f(k_1,h_1)f(k_2,h_2)$$

$\textbf{Question:}$ Is the above correct? And how to prove the "only if" part?

Answer 1

Well, in the group $H\times K$ the subgroups $T=H\times \{e\},R=\{e\}\times K$ are normal (it is straightworward check), and also $T\cap R=\{e\}$ and $TR=H\times K$, so, if f is an isomorphism, then the above properties also hold for H, K and G (direct check).

Answer 2

There's a slicker proof to show that if $H$ and $K$ are both normal and $H\cap K=\{1\}$ then $hk=kh$ for all $h\in H$, $k\in K$: consider the element $hkh^{-1}k^{-1}$. You can write it both as $$h(kh^{-1}k^{-1})\in H$$

$$(hkh^{-1})k^{-1}\in K$$

So $hkh^{-1}k^{-1}\in H\cap K=\{1\}$ and $hk=kh$.

The rest of your proof is correct as well, this can just make it a little nicer