A question about natural epimorphism

Question

My question came from a proof of a proposition about perfect groups, it quote it here:

Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect.

At the beginning of the proof, it writes that

From the fact that $[x,y]^\varphi=[x^\varphi,y^\varphi]\Longrightarrow (G')^\varphi=(G^\varphi)'$, applied to the natural epimorphism, we obtain $$G/N={\color{red}{(G/N)'=G'N/N}}.$$

I think the $\varphi$ here can refer to the natural epimorphism; however, I think it should be ${\color{red}{G'/N}}$, not ${\color{red}{G'N/N}}$, but in that way, since $N$ is not necessarily contained in $G'$, it will contradict the requirement of natural epimorphism that $N\trianglelefteq G'$. Both sides seem unreasonable to me.

I'm now completely confused, how can it be $G'N/N$?

Is my question clear? Any help will be sincerely appreciated!

PS: It’s on page 25 of my textbook.

Answer 1

The magic words are "Second Isomorphism Theorem".

Answer 2

When you write $G'/N$, you must make sure $N\unlhd G'$. So you must have $N\subset G'$. But here we don't necessarily have that.

Let $\varphi:G\to G/N$ with $\varphi(g)=gN$.

Claim: $\varphi(G')=G'N/N$.

Proof. Since $N\unlhd G'N$, $G'N/N$ is well defined. Let $g\in G'$, then $\varphi(g)=gN\in G'N/N$. Conversely, let $gnN\in G'N/N$. Then $gnN=gN=\varphi(g)\in \varphi(G')$.

As mentioned in the comments above, this is in 1.2.6 in your textbook.