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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.

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2 Answers 2

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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}

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    $\begingroup$ 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/… $\endgroup$ Nov 20, 2020 at 1:40
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Hint. For $h ∈ H$, $k ∈ K$, you have $hk = kh \iff hkh^{-1}k^{-1} = 1$.

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  • $\begingroup$ 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! $\endgroup$ Nov 19, 2020 at 7:12

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