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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

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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
@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
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
@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
up vote 2 down vote accepted

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

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

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

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

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