Quotient group by normal closure of union

Question

Let $G$ be a group and $H, N \subseteq G$ be subsets of the group. Let $\overline{A}$ denote the normal closure of any subset $A\subseteq G'$ in some group $G'$. Let $\pi: G \to G/\overline{N}$ denote the quotient map.

I want to know if it is true that the following groups are isomorphic $$G/\overline{N\cup H} \cong (G/\overline{N})/\overline{\pi(H)}$$

I think I have to show that $\overline{\pi(H)}=\pi(\overline{N \cup H})$ so that I can use isomorphism theorems.

So, to show the inclusion $\overline{\pi(H)}\subseteq \pi(\overline{N \cup H})$ note that $\pi(H)\subseteq \pi(N \cup H) \subseteq \pi( \overline{N \cup H})$, where the latter is a normal subgroup of $G/\overline{N}$. But then by the definition of normal closure $\overline{\pi(H)}\subseteq \pi(\overline{N\cup H})$.

To show the other inclusion, I want to show that $N \cup H \subseteq \pi^{-1}(\overline{\pi(H)})$. Since $\overline{\pi(H)}$ is a subgroup it contains the identity element. But then it's inverse under $\pi$ contains the kernel of $\pi$ which contains $N$. We also have that $H \subseteq \pi^{-1}(\pi(H))\subseteq \pi^{-1}(\overline{\pi(H)})$. Since the preimage of a normal subgroup under a homomorphism is a normal subgroup we have that $N \cup H \subseteq \pi^{-1}(\overline{\pi(H)})$ implies that $\overline{N\cup H} \subseteq \pi^{-1}(\overline{\pi(H)})$. Now since $\pi$ is surjective, we conclude $\pi(\overline{N\cup H}) \subseteq \overline{ \pi(H)}$.

Now we can use the third isomorphism theorem, since $\pi(\overline{N \cup H})=\overline{N\cup H}/\overline{N}$, because $\overline{N}$ is a normal subgroup of $\overline{N \cup H}$:

$$ (G/\overline{N})/\overline{\pi(H)} = (G/\overline{N})/\pi(\overline{N \cup H}) = (G/\overline{N})/(\overline{N \cup H}/\overline{N}) \cong G/\overline{N\cup H}$$.

Am I reasoning correctly, is there are simpler proof of this?

Answer 1

Let $\sigma:G/\overline{N}\to(G/\overline{N})/\overline{\pi(H)}$ be the natural quotient.

What is $K=\text{Ker}(\sigma\circ\pi)$? It is the smallest normal subgroup of $G$ that contains $h\overline{N}$ for all $h\in H$.

We just need to show that $K=\overline{N\cup H}$.

• $\overline{N\cup H}$ is a normal subgroup of $G$. Furthermore, $\overline{N\cup H}\supseteq \overline{N}\;$ and $\;\overline{N\cup H}\supseteq h$ for all $h\in H$. Therefore, because a subgroup is closed under multiplication, $\overline{N\cup H}\supseteq h\overline{N}$ for all $h\in H$. It follows that $\overline{N\cup H}\supseteq K$.
• Since $en=n$ and $he=h$ are elements of $K$ for all $n\in N$ and $h\in H$, we have that $K$ contains $N\cup H$. Since $K$ is a normal subgroup of $G$, we have that $K$ contains $\overline{N\cup H}$.