Direct product and commutativity of factors

Question

If $H$ and $K$ are both normal subgroups of $G$, and if every element of $G$ can be written uniquely as $hk$ with $h\in H$ and $k\in K$ (so $G$ is the direct product of $H$ and $K$), does it follow that $H$ and $K$ commute, i.e. that $kh=hk$ always?

Answer 1

Yes, that is always true. Note that $G$ being the direct product of $H$ and $K$ means that the map $\Phi\colon G\to H\times K$ given by $hk\mapsto (h,k)$ is an isomorphism, where $H\times K$ carries the group structure given by $(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$. Under this isomorphism, note that $$ \Phi(kh) = \Phi((1k)(h1)) = \Phi(1k)\Phi(h1) = (1,k)(h,1) = (1h,k1) = (h,k) = \Phi(hk). $$ So $\Phi(kh)=\Phi(hk)$ and hence $kh=hk$.

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To elaborate on the direct proof suggested by @AugustLiu in the comments: We have $H\cap K = \{1\}$, since otherwise $1\neq g\in H\cap K$ would have the two different expressions $$ g = \underbrace{g}_{\in H} \cdot \underbrace{1}_{\in K} = \underbrace{1}_{\in H} \cdot \underbrace{g}_{\in K}. $$ Now note that since $H$ and $K$ are normal subgroups, we have $$ H \ni \underbrace{(khk^{-1})}_{\in H} \, \underbrace{h^{-1}}_{\in H} = \underbrace{k}_{\in K} \underbrace{(hk^{-1}h^{-1})}_{\in K} \in K $$ so that $khk^{-1}h^{-1}\in H\cap K=\{1\}$. Hence $khk^{-1}h^{-1}=1$, which is equivalent to $kh=hk$.

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Yet another proof goes like this: The element $kh\in G$ must have a unique expression $kh=h'k'$ with $h'\in H$ and $k'\in K$. However, this yields $$ h = k^{-1} h' k' = \underbrace{(k^{-1} h' k)}_{\in H} \underbrace{(k^{-1} k')}_{\in K} $$ and the unique expression of $h$ as a product of an element in $H$ and an element in $K$ is $h=h\cdot 1$, so that $1=k^{-1} k'$ and hence $k'=k$. By a similar argument $h'=h$ and hence $kh=hk$.