# $H$ and $K$ are two normal subgroups of group $G$, $H\cap K=\{1\}$, $G=HK$, how to prove $G$ is isomorphic to $H\times K$

If group $$G$$ has two normal subgroups $$H$$ and $$K$$ satisfying $$H\cap K=\{1\}$$ and $$G=HK$$, how to prove that $$G$$ is isomorphic to $$H\times K$$?

Since the factoring of any $$g\in G$$ as product $$hk$$ is unique under given conditions, I tried to prove that the map $$\phi:G\ni g\mapsto (h,k)\in H\times K$$ is the isomorphism. I can show that $$\phi$$ is a bijection. Next, I need to show it preserves group operation. Assuming $$\forall g_1,g_2\in G$$, $$g_1=h_1k_1$$ and $$g_2=h_2k_2$$ (the factoring is unique), we have $$g_1g_2=h_1k_1h_2k_2$$. But $$\phi(g_1)=(h_1,k_1),\phi(g_2)=(h_2,k_2)$$, so in product group $$H\times K$$, the operation $$\phi(g_1)\phi(g_2)=(h_1h_2,k_1k_2)$$. Its pre-image of $$\phi$$ in $$G$$ should be the product $$h_1h_2k_1k_2$$. So we must establish $$h_1k_1h_2k_2=h_1h_2k_1k_2$$, which is equivalent to $$k_1h_2=h_2k_1$$. But I cannot prove this commutative relation based on the given conditions. The normality of $$H$$ implies $$k_1^{-1}h_2k_1\in H$$, but it does not necessarily equal $$h_2$$. This is where I got stuck. Is the above commutative relation true and how do I prove it? Or, was I wrong at the very beginning about what the isomorphism is? Could you please give me some help? Thank you.

This doesn't add much to your excellent effort, other than some details.

For any pair of subgroups $$H, K\le G$$, the set $$H\times K$$ can be partitioned into equivalence classes all equicardinal to $$H\cap K$$, and such that $$(H\times K)/\sim$$ is equicardinal to $$HK$$. In fact:

Let's define in $$H\times K$$ the equivalence relation: $$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$$. The equivalence class of $$(h,k)$$ is given by:

$$[(h,k)]_\sim=\{(h',k')\in H\times K\mid h'k'=hk\} \tag 1$$

Now define the following map from any equivalence class:

\begin{alignat*}{1} f_{(h,k)}:[(h,k)]_\sim &\longrightarrow& H\cap K \\ (h',k')&\longmapsto& f_{(h,k)}((h',k')):=k'k^{-1} \\ \tag 2 \end{alignat*}

Note that $$k'k^{-1}\in K$$ by closure of $$K$$, and $$k'k^{-1}\in H$$ because $$k'k^{-1}=h'^{-1}h$$ (being $$(h',k')\in [(h,k)]_\sim$$) and by closure of $$H$$. Therefore, indeed $$k'k^{-1}\in H\cap K$$.

Lemma 1. $$f_{(h,k)}$$ is bijective.

Proof.

\begin{alignat}{2} f_{(h,k)}((h',k'))=f_{(h,k)}((h'',k'')) &\space\space\space\Longrightarrow &&k'k^{-1}=k''k^{-1} \\ &\space\space\space\Longrightarrow &&k'=k'' \\ &\stackrel{h'k'=h''k''}{\Longrightarrow} &&h'=h'' \\ &\space\space\space\Longrightarrow &&(h',k')=(h'',k'') \\ \end{alignat}

and the map is injective. Then, for every $$a\in H\cap K$$, we get $$ak\in K$$ and $$a=f_{(h,k)}((h',ak))$$, and the map is surjective. $$\space\space\Box$$

Now define the following map from the quotient set:

\begin{alignat}{1} f:(H\times K)/\sim &\longrightarrow& HK \\ [(h,k)]_\sim &\longmapsto& f([(h,k)]_\sim):=hk \\ \tag 3 \end{alignat}

Lemma 2. $$f$$ is well-defined and bijective.

Proof.

• Good definition: $$(h',k')\in [(h,k)]_\sim \Rightarrow f([(h',k')]_\sim)=h'k'=hk=f([(h,k)]_\sim)$$;
• Injectivity: $$f([(h',k')]_\sim)=f([(h,k)]_\sim) \Rightarrow h'k'=hk \Rightarrow (h',k')\in [(h,k)]_\sim \Rightarrow [(h',k')]_\sim=[(h,k)]_\sim$$;
• Surjectivity: for every $$ab\in HK$$ , we get $$ab=f([(a,b)]_\sim)$$. $$\space\space\Box$$

As a corollary, if $$|H\cap K|=1$$, then all the classes $$[(h,k)]_\sim$$ are singletons, and hence the map $$\tilde f\colon H\times K\to (H\times K)/\sim$$, defined by $$(h,k)\mapsto [(h,k)]_\sim$$, is bijective; this, in turn, implies that the composite map $$f\circ\tilde f\colon H\times K\to HK$$ is bijective, too. If, in addition, $$H\unlhd G$$ and $$K\unlhd G$$, then for every $$h\in H, k\in K$$ we have that $$hkh^{-1}k^{-1}\in H\cap K$$, whence $$hkh^{-1}k^{-1}=e$$ and finally $$hk=kh$$. This, and the assumption $$HK=G$$, make of $$f\circ\tilde f$$ a (bijective) group homomorphism (namely an isomorphism). In fact:

\begin{alignat}{1} (f\circ\tilde f)((h,k)(h',k')) &= (f\circ\tilde f)(hh',kk') \\ &= f(\tilde f(hh',kk')) \\ &= f([hh',kk']_\sim) \\ &= hh'kk' \\ &= hkh'k' \\ &= (hk)(h'k') \\ &= f([h,k]_\sim)f([h',k']_\sim) \\ &= f(\tilde f(h,k))f(\tilde f(h',k')) \\ &= ((f\circ\tilde f)(h,k))((f\circ\tilde f)(h',k')) \\ \end{alignat}

• Thank you. This answer gives me a deeper and more direct understanding of the result. The isomorphism $H\times K\cong\rm{(when\;H\cap K=\{1\})}\frac{H\times K}{H\cap K \rm{(\cong Stab\;1)}}\cong HK=G$ can be proved elegantly by group action: math.stackexchange.com/questions/168942/… Nov 20, 2020 at 1:40

Hint. For $$h ∈ H$$, $$k ∈ K$$, you have $$hk = kh \iff hkh^{-1}k^{-1} = 1$$.

• I see, $hkh^{-1}k^{-1}=(hkh^{-1})k^{-1}=h(kh^{-1}k^{-1})$, so it is in both $H$ and $K$. Thank you! Nov 19, 2020 at 7:12