Suppose that $H$ and $K$ are normal subgroups of $G$ such that $H \cap K=\{e\}$. Show that $G$ is isomorphic with some subgroup of $G/H \times G/K$.

Question

Let $G$ be a group with identity $e$. Suppose that $H$ and $K$ are normal subgroups of $G$ such that $H \cap K=\{e\}$. Show that $G$ is isomorphic with some subgroup of $G/H \times G/K$.

It seems that from $H \cap K=\{e\}$ I have that $hk=kh$ for all $h\in H, k\in K$. I don’t know if I can apply any of the isomorphism theorems here since I don’t know if $H \subset K$ or if $K \subset H$. Hints on what can we do here?

Answer 1

Since $H,K\lhd G$, so $\forall g\in G, gH=Hg, gK=Kg$.

Let the map $\varphi:G\to\frac GH\times\frac GK$ as $\varphi(g)=(gH,gK)$.

Part 1: Show this map is injection

If $\varphi(x)=\varphi(y)$, then $xH=yH, xK=yK$. This is equivalent to $y^{-1}xH=H, y^{-1}xK=K$.

This means $y^{-1}x\in H,K$, so $y^{-1}x\in H\cap K$,so $y^{-1}x$ must be $e$, so $x=y$.

So, $\varphi$ is injection.

Part 2: Show this map is homomorphism

Then, let's calculate $\varphi(x)\varphi(y)$. $$\varphi(x)\varphi(y)=(xH,xK)\cdot(yH,yK)=(xHyH,xKyK)=(xyHH,xyKK)=(xyH,xyK)=\varphi(xy)$$

So, $\varphi$ is group homomorphism.

Last part: Prove the question

Then, $\varphi:G\to\varphi G$ is isomorphism($\varphi$ in this case is bijective homomorphism.), and $\varphi G$ is subgroup of $\frac GH\times\frac GK$.