# If $H$ and $K$ are normal subgroups of $G$ and $H\bigcap K = \{e\}$, prove that $G$ is isomorphic to a subgroup of $G/H \times G/K$

I tried proving this in the following manner, but I am not confident with these types of problems so any verification would be appreciated. Thank you.

Let $$A = \{(gH, gK): g \in G\}$$

Define $$\phi$$ : $$G$$ $$\rightarrow$$ $$A$$ by $$\phi(g)=(gH,gK)$$

First we'll show $$\phi$$ is a homomorphism:

$$\phi(gg')=(gg'H,gg'H)=(gHg'H,gHg'H)=(gH,gH)(g'H,g'H)=\phi(g)\phi(g')$$

Now we'll show $$\phi$$ is bijective, and thus an isomorphism:

The codomain of $$\phi$$ is {$$(gH,gK):g \in G$$} so $$\phi$$ is clearly onto. Now suppose $$g_1\not= g_2$$ and $$\phi(g_1)=\phi(g_2)$$. Then $$(g_1H,g_1K)=(g_2H,g_2K)$$ $$\Rightarrow$$ $$g_1H=g_2H$$ and $$g_1K=g_2K$$ $$\Rightarrow$$ $$g_2^{-1}g_1\in H$$ and $$g_2^{-1}g_1 \in K$$, a contradiction since we assumed $$H\bigcap K = \{e\}$$

$$\square$$

• Your argument is sound. You could also prove injectivity directly, instead of by contradiction. Oct 21, 2018 at 6:43
• Ah so you would just say $g_2^{-1}g_1 \not= e$ $\Rightarrow$ $g_2^{-1}g_1 \notin H \bigcap K$ $\Rightarrow$ either $g_1H \not= g_2H$ or $g_1K \not= g_2K$? Oct 21, 2018 at 7:10
• I would say $g_2^{-1}g_1 \in H \cap K = \{e \} \implies g_2^{-1} g_1 = e \implies g_1 = g_2$. Oct 21, 2018 at 7:18

\begin{align*} \ker\phi &=\{g\in G\mid (gH,gK)=(H,K)\}\\ &= \{g\in G\mid gH=H\text{ and }gK= K\}\\ &= \{g\in G\mid g\in H\text{ and }g\in K\}\\ &= H\cap K\\ &= \{e\}\end{align*}