Suppose $H$ and $K$ are normal subgroups of a group $G$. Show that there exists a 1-1 homomorphism $\phi:G/(H\cap K)\rightarrow G/H \times G/K$.
Here are my thoughts:
If we can show that $\ker\phi=1$ then we have that $\phi$ is 1-1. So that's what I'm trying to do.
$G/H\cap K=\{gH\mid g\in K\}$
$G/H\times G/K=\{(gH,gK)\mid g\in G\}$
Then I considered the mapping where $\phi(gH)=(gH,gK)$ and looked at the kernel.
$\ker\phi$ $$=\{g\in G\mid (gH, gK)=(H,K)\}$$ $$=\{g\in G\mid g\in H\text{ and }g\in K\}$$ $$=H\cap K.$$
However, now I'm stuck. I'm not sure how to show that the intersection is just the identity. Am I on the right track? Any help is appreciated.