Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $H$ and $K$ are normal subgroups of a group $G$ such that $HK=G$. I need to show that $G/(H\cap K)\cong (G/H)\times (G/K)$.

So from the second isomorphism theorem we have that:

$HK/H\cong H/(H\cap K)$ which gives that $G/H\cong H/(H\cap K)$

$HK/K\cong K/(H\cap K)$ which gives that $G/K\cong K/(H\cap K)$

So we must now show that:

$H/(H\cap K)\times K/(H\cap K)\cong G/(H\cap K)$

So consider the map $\theta:H/(H\cap K)\times K/(H\cap K)\rightarrow G/(H\cap K)$ such that $\theta(h(H\cap K)\times k(H\cap K))=hk(H\cap K)$

Now this is a homomorphism and is surjective. To check injectiviy we see that if:

$\theta(h_1(H\cap K)\times k_1(H\cap K))=\theta(h_2(H\cap K)\times k_2(H\cap K))$ then we have that:

$(h_1k_1(H\cap K))=(h_2k_2(H\cap K))$ so that $(h_1k_1)(h_2k_2)^{-1}(H\cap K)=(H\cap K)$ which gives that $h_1k_1(H\cap K)=h_2k_2(H\cap K)$ So we are done.

Is the above correct?

Thanks for any help

share|improve this question
1  
I might be wrong: Let $a=(h_1(H\cap K),k_1(H\cap K)),b=(h_2(H\cap K),k_2(H\cap K))$ where $a,b\in H/(H\cap K)\times K/(H\cap K)$. Then $ab=(h_1h_2(H\cap K),k_1k_2(H\cap K)$. But $\theta(ab)=h_1h_2k_1k_2$ while $\theta(a)\theta(b)=h_1k_1h_2k_2$ so $\theta(ab)\neq \theta(a)\theta(b)$. It is equal when $k_1h_2=h_2k_1$, i.e $K,H$ commute. This means $K\cap H=\lbrace e\rbrace$, but for the question it is not necessarily so. –  Yong Hao Ng Dec 9 '12 at 11:10
    
Haha yeah, that was a bit silly. I was thinking that as $H,K$ are normal in $G$ then we can use that but of course that only gives $h_1h_2k'k_2$ for some $k'\in K$. –  hmmmm Dec 9 '12 at 11:21
1  
@YongHaoNg Do we not only need that $H/(H\cap K)$ and $K/(H\cap K)$ intersect trivially? –  hmmmm Dec 9 '12 at 11:44
1  
You have interchanged the two expressions in the second isomorphism theorem, and the "So we are done" at the end hides a final step (which would be shorter anyway if you regarded $\theta: H \times K \to G/(H\cap K)$ instead). Apart from that, it is fine. –  Phira Dec 9 '12 at 12:31
1  
@hmmmm I guess I am wrong after all. You are right, $H/(H\cap K)$ and $K/(H\cap K)$ intersect trivially by construction (common element is in both $H$ and $K$). I did not put in $(H\cap K)$ for $h_1h_2k_1k_2(H\cap K)$ and $h_1k_1h_2k_2(H\cap K)$. Otherwise intersecting trivially means they commute $(k_1h_2=h_2k_1$ or otherwise it is in $H\cap K$). –  Yong Hao Ng Dec 9 '12 at 13:46

1 Answer 1

up vote 2 down vote accepted

To show $\,\theta\,$ is injective it's simpler, imo, to consider its kernel:

$$x\in H\,,\,y\in K\,\;\;:\;\;\,(x(H\cap K),y(H\cap K))\in\ker\theta\Longrightarrow$$

$$\Longrightarrow xy\in H\cap K \Longrightarrow\begin{cases}xy=h\in H\Longrightarrow y=x^{-1}h\in H\\xy=k\in K\Longrightarrow x=ky^{-1}\in K\end{cases}\Longrightarrow$$

$$\Longrightarrow x\in H\cap K\,,\,y\in H\cap K\Longrightarrow,(x(H\cap K),y(H\cap K))=(1,1)\Longrightarrow \ker\theta=1$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.