# $H$ 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.

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 '20 at 7:12

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 '20 at 1:40