Isomorphism (Help)

**Markeur** · November 3rd 2010, 07:42 AM

If $H$ and K are normal subgroups of a group $G$ with $HK=G$ , prove that $G/(H \cap K) \simeq (G/H) \times (G/K)$ .
Hint: If $\phi:G \rightarrow (G/H) \times (G/K)$ is defined by $x \mapsto (xH,xK)$ , then $ker \: \phi=H \cap K$ ; moreover, we have $G=HK$ , so that $\displaystyle\cap_{a}aH=HK=\displaystyle\cap_{b}bK$ .

I have managed to prove that $ker \: \phi=H \cap K$ . I'm having difficulty in proving that $\phi$ is surjective. This is crucial in proving the above isomorphism using 1st Isomorphism Theorem.

**Drexel28** · November 3rd 2010, 08:14 AM

Originally Posted by Markeur

If $H$ and K are normal subgroups of a group $G$ with $HK=G$ , prove that $G/(H \cap K) \simeq (G/H) \times (G/K)$ .
Hint: If $\phi:G \rightarrow (G/H) \times (G/K)$ is defined by $x \mapsto (xH,xK)$ , then $ker \: \phi=H \cap K$ ; moreover, we have $G=HK$ , so that $\displaystyle\cap_{a}aH=HK=\displaystyle\cap_{b}bK$ .

I have managed to prove that $ker \: \phi=H \cap K$ . I'm having difficulty in proving that $\phi$ is surjective. This is crucial in proving the above isomorphism using 1st Isomorphism Theorem.

Let $\left(yH,zK\right)$ be in the codomain. Since $HK=G$ what can you say about $y,z$ ?

**Markeur** · November 3rd 2010, 09:30 AM

Hey,

The proof is as below:

We want to show that $\forall (yH,zK) \in (G/H) \times (G/K)$ , $\exists g \in G$ such that $\phi(g)=(gH,gK)$ .

Since $(yH,zK) \in (G/H) \times (G/K)$ , $yH \in G/H$ and $zK \in G/K$ .
Since $G=HK$ , $yH \in HK/H$ and $zK \in HK/K$ .
This implies that $y,z \in HK$ .
Since $G=HK$ , $g \in HK$ .
This implies that $y=g$ and $z=g \: \forall y,z \in G$ .
Hence $\exists g \in G$ such that $\phi(g)=(gH,gK)$ .
Therefore $(G/H) \times (G/K) \subseteq Im \; \phi$ .
Clearly, $Im \; \phi \subseteq (G/H) \times (G/K)$ .
Finally, we conclude that $(G/H) \times (G/K)=Im \; \phi$ .

From there, we can establish the isomorphism using First Isomorphism Theorem.

So, is the above proof correct?

Thanks in advance.

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