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

Question

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 $H\bigcap K = \{e\}$

$\square$

Answer 1

You could make this cleaner by just directly calculating the kernel (though like somebody already said, your argument is valid):

\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*}