Q: Please help me with proving a subgroup is normal Question:

Let $G$ be a group and $N = \{g_1h_1g_1^{-1}h_1^{-1} \dots g_nh_ng_n^{-1}h_n^{-1} \mid n \in \Bbb N, g_i, h_i \in G\}$. Show that $N$ is a normal subgroup of $G$.

I have shown that $N$ is a subgroup of $G$, but I can't show that it is normal. The only tool I can think of is the definition, so I pick arbitrary $g_0$ from $G$ and put it and its inverse on both side of an element of $N$, but I can't see why this new thing is in $N$. I tried to write $g_0 = ghg^{-1}h^{-1}$ for some $g, h \in G$ but clearly this does not work in Abelian case. (Perhaps this will work in non-Abelian case? We don't need to care about Abelian groups since their subgroups are normal.)

A: Your approach of conjugating an arbitrary element of $N$ is fundamentally sound, but you need to elaborate the argument to rewrite the conjugates of commutators as commutators themselves.
Observe that you can rewrite
$$
\begin{align*}
g_0(g_1h_1g_1^{-1}h_1^{-1})g_0^{-1}
& = 
(g_0g_1g_0^{-1})(g_0h_1g_0^{-1})(g_0g_1^{-1}g_0^{-1})(g_0h_1^{-1}g_0^{-1})\\
& = (g_0g_1g_0^{-1})(g_0h_1g_0^{-1})(g_0g_1g_0^{-1})^{-1}(g_0h_1g_0^{-1})^{-1}
\end{align*}
$$
I leave it to you to argue either the general case, or why this implies the general case.  (Note that a slight variation of this argument shows that $N$ is not just normal but fully characteristic.)
