$\frac{G_1\times G_2}{N_1\times N_2}\cong \bigg(\frac{G_1}{N_1}\bigg)\times \bigg(\frac{G_2}{N_2}\bigg)$

Question

Is my solution to this question correct?

If $N_1\triangleleft G_1,N_2\triangleleft G_2$, then $(N_1\times > N_2)\triangleleft (G_1\times G_2)$ and $\frac{G_1\times G_2}{N_1\times N_2}\cong \bigg(\frac{G_1}{N_1}\bigg)\times \bigg(\frac{G_2}{N_2}\bigg)$

First part

$(g_1,g_2)\in G_1\times G_2$ and $(n_1,n_2)\in N_1\times N_2$, then $(g_1,g_2)(n_1,n_2)(g_1^{-1},g_2^{-1})=(g_1n_1g^{-1}_1,g_2n_2g^{-1}_2)\in N_1\times N_2$, since $N_1$ and $N_2$ are normal subgroups of $G_1$ and $G_2$ respectively.

Second part

Let $\varphi: G_1\times G_2\to \frac{G_1}{N_1}\times\frac{G_2}{N_2}$ be the canonical epimorphism given by $(a,b)\mapsto(a+N_1,b+N_2)$. Note that $\ker \varphi =N_1\times N_2$, because obviously $N_1\times N_2\subset \ker\varphi$ and let $(a,b)\in \ker\varphi\implies\varphi(a,b)=(a+N_2,b+N_2)=(0+N_1,0+N_2)\implies a\in N_1$ and $b\in N_2\implies (a,b)\in N_1\times N_2$.

Thus by first isomorphism theorem we have $\frac{G_1\times G_2}{N_1\times N_2}\cong \frac{G_1}{N_1}\times \frac{G_2}{N_2}$.

Thanks in advance

Answer 1

Yes, this is a correct solution. But you can leave out the first part, because if $N_1\times N_2$ is the kernel of a homomorphism, it is normal. The kernel of a homomorphism is always normal.