Suppose $H$ and $K$ are normal subgroups of $G$. Prove that $G/H \times G/K$ has a subgroup that is isomorphic to $G / (H∩K)$.

Question

Suppose $H$ and $K$ are normal subgroups of $G$. Prove that $G/H \times G/K$ has a subgroup that is isomorphic to $G / (H∩K)$.

Also prove that if $G = HK$, then $G/(H∩K)$ is isomorphic to $G/H \times G/K$

So far I have that by the second isomorphism theorem, $H/(H ∩ K)$ is isomorphic to $HK/K$. I'm not sure where to go from here.

Answer 1

Hint: there is a natural homomorphism $$ \varphi\colon G\to G/H\times G/K $$ defined by $\varphi(x)=(xH,xK)$. What's its kernel? What's its image in case $HK=G$?

Answer 2

Define

$$\phi: G\to G/H\times G/K\;,\;\;\phi x:=(xH,\,xK)$$

Prove the above is a group homomorphism and show $\;\ker\phi=H\cap K\;$ . Finally, use the first isomorphism theorem.

If you understand the above then the second part should be easier.