I am given a group $H$ and two normal subgroups $A$ and $B$ and $H/A$ and $H/B$ are two quotient groups. A homomorphism is defined as follows: $\phi: H \to H/A \times H/B; \phi(h) = (hA,hB)$. How do I show that the Ker($\phi$) = $A \cap B$? So I know, that the identity element in the co-domain is simply $(A,B)$.
Attempt: If I take $m \in A \cap B$, then by normality of $A \cap B$, I know that $hmh^{-1} \in A \cap B$. Then $\phi(hmh^{-1}) = \phi(h) \phi(m) \phi(m^{-1}) = (hA,hB) \circ (mA, mB) \circ (h^{-1}A, h^{-1}B) =(hmh^{-1}A, hmh^{-1}B)$
How do I proceed? I am not sure if I am heading towards the right direction. Can someone help?
If $m\in A\cap B$, then $mA=A$ and $mB=B$ and so $\phi(m)=(mA, mB)=(A, B)$ which is the identity element of $H/A\times H/B$. So $m\in Ker\phi$.