Prove $(G_1 \times G_2)/(N_1 \times N_2) \cong G_1/N_1 \times G_2/N_2$.

Problem Statement

Suppose $G_1, G_2$ are groups with $N_1 \lhd G_1$ and $N_2 \lhd G_2$ normal subgroups. Prove that $N_1 \times N_2 \lhd G_1 \times G_2$ and $(G_1 \times G_2)/(N_1 \times N_2) \cong G_1/N_1 \times G_2/N_2$.

Preamble

For this, I have a proof outlined and would like feedback on whether it's correct and where it could be simplified - all that good stuff. I will not include the proof that $N_1 \times N_2 \lhd G_1 \times G_2$ since that's a straightforward application of the definition of a normal subgroup as well as the computation of the direct product.

The proof of the second statement is, I believe, a case of the First Isomorphism Theorem. To that end, I endeavored to find a surjective homomorphism from $G_1 \times G_2$ to $G_1/N_1 \times G_2/N_2$ whose kernel was the normal subgroup $N_1 \times N_2$. I believe my proof is correct though my grasp of the material, specifically operating with quotient groups, is less intuitive than I would like. The possibility that I've made some oversight or mistake in my logic is a very real one. Due to this lack of intuition the proof will include a bit of my thought process as well, so it will be longer than is necessary but I want to include the thinking so that if something is wrong it can be clear where that misinterpretation is coming from. In particular my proof that the kernel of the homomorphism is the normal subgroup will include this extended thought process.

The Proof

Let $\eta: G_1 \times G_2 \to G_1/N_1 \times G_2/N_2$ be defined by $$\eta((g_1,g_2)) = (g_1N_1, g_2N_2)$$

1. Now this mapping is a homomorphism since for any $(g_1, g_2),(h_1, h_2) \in G_1 \times G_2$: $$\eta((g_1, g_2)(h_1, h_2)) = ((g_1h_1)N_1, (g_2h_2)N_2) = ((g_1N_1)(h_1N_1), (g_2N_2)(h_2N_2)) = \eta((g_1,g_2))\eta((h_1,h_2))$$ Where the second equality holds since the product of two left cosets is itself a left coset.

2. $\eta$ is surjective since for any $(g_1N_1, g_2N_2) \in G_1/N_1 \times G_2/N_2$ there exists a $(g_1,g_2) \in G_1 \times G_2$ such that $\eta((g_1,g_2)) = (G_1N_1,g_2N_2)$

3. The kernel of $\eta$ is a subgroup of $G_1 \times G_2$ such that all elements in the kernel get sent to the identity in $G_1/N_1 \times G_2/N_2$.

   • The identity in $G_1/N_1 \times G_2/N_2$ is $N_1 \times N_2 = (n_1 , n_2)\ \text{for any}\ n_1,n_2 \in N_1 \times N_2$ since $$(g_1N_1,g_2N_2)(n_1,n_2) = (g_1N_1n_1, g_2N_2n_2) = (g_1N_1, g_2N_2)$$ Since $N_1,N_2$ are subgroups and hence closed under products and since all right identities are left identities we can see $N_1 \times N_2$ is the identity.
   • Now we can take any $(n_1,n_2) \in N_1 \times N_2$ and see that $$\eta((n_1,n_2)) = (n_1N_1,n_2N_2) = (N_1,N_2)$$ Since $(n_1,n_2)$ is an arbitrary element of the group $N_1 \times N_2$ we can see that $ker(\eta) = N_1 \times N_2$.

Thus, we can appeal to the First Isomorphism Theorem to see that we are guaranteed an isomorphism between $(G_1 \times G_2)/(N_1 \times N_2)$ and $G_1/N_1 \times G_2/N_2$ and hence $(G_1 \times G_2)/(N_1 \times N_2) \cong G_1/N_1 \times G_2/N_2. \square$