let $N_1$, $N_2$ be normal subgroups of group G. Define $f : G → (G/N_1) × (G/N_2)$ by $f(a) = (aN_1, aN_2)$. show f is a homomorfism Let G be a group, and let N_1, N_2 be normal subgroups of G. Define $f : G → (G/N_1) × (G/N_2)$ by $f(a) = (aN_1, aN_2)$.
(a) Prove that f is a group homomorphism with kernel N1 ∩ N2.
(b) Prove that G/(N1 ∩ N2) is isomorphic to a subgroup of (G/N1) × (G/N2).
I'm already stuck at showing this is a homomorphism. How can i show $(abN_1, abN_2)=(aN_1, aN_2)(bN_1, bN_2)$?
 A: Well, $(aN_1)(bN_1)$ is defined to be $abN_1$ and $(x, y)(z, w)$ is defined to be $(xz, yw)$ and so $(aN_1, aN_2)(bN_1, bN_2)=((aN_1)(bN_1), (aN_2)(bN_2))=(abN_1, abN_2)$.
A: Since someone already helped you prove $f$ is a homomorphism, I'll help you with the rest.
If $a \in \text{ker}(f)$, then $f(a) = (aN_1, aN_2) = (eN_1, eN_2)$, because with the definition of the operation on the quotient group, it follows that $eN_1$ is the identity element of $G/N_1$ and $eN_2$ is the identity element of $G/N_2$.
If $aN_1 = eN_1$, then for some $n_1, n_2 \in N_1$ have $an_1 = en_2$, so $a = n_2n_1^{-1} \in N_1$. More generally, one can show analogously that $aH = bH \iff b^{-1}a \in H$, for any subgroup $H$ of a group $G$.
So we have $a \in N_1$, and since we also have $aN_2 = eN_2$, we also get $a \in N_2$, so then $a \in N_1 \cap N_2$. This proves $\text{ker}(f) \subseteq N_1 \cap N_2$.
The other inclusion, $N_1 \cap N_2 \subseteq \text{ker}(f)$, follows from the other implication from theorem I just stated, meaning the implication $b^{-1}a \in H \implies aH = bH$. If you want, try and prove this theorem yourself for an additional exercise (hint: write $b^{-1}a = h \in H$ and use this to rewrite elements of $aH$ and $bH$).
For question b, you should look up the first isomorphism theorem and understand how it works, if you haven't already. If you do that you should be able to work things out.
