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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.

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  • $\begingroup$ I'm trying to fathom the motivation to write the application of a homomorphism "exponentially"... $\endgroup$ – Derek Elkins left SE Mar 6 '18 at 3:48
  • $\begingroup$ @DerekElkins Whichever is ok : ) $\endgroup$ – Hello Mar 6 '18 at 3:50
  • $\begingroup$ @DerekElkins I am quite sure that the exponential notation is for conjugation (inner automorphism). $\endgroup$ – Igor Rivin Mar 6 '18 at 3:59
  • $\begingroup$ @DerekElkins $a^\varphi=\varphi(a)$, where $\varphi$ is a homomorphism. $\endgroup$ – Hello Mar 6 '18 at 4:02
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    $\begingroup$ @IgorRivin Then you are quite wrong (in this case). Look at section 1.2 of the referenced PDF. $\endgroup$ – Derek Elkins left SE Mar 6 '18 at 4:10
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The magic words are "Second Isomorphism Theorem".

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  • $\begingroup$ THANKS! I know the theorem. How can it help me? The theorem is applied later in the proof, so this theorem might not be the key here to my question. What have I missed? $\endgroup$ – Hello Mar 6 '18 at 3:57
  • $\begingroup$ Look at the statement of the theorem, which tells you that $G^\prime N/N = G^\prime/(G^\prime \cap N).$ So, you need to check that $(G/N)^\prime = G^\prime/(G^\prime \cap N).$ But the RHS is just $\phi(G^\prime),$ while the LHS is $(\phi(G))^\prime.$ $\endgroup$ – Igor Rivin Mar 6 '18 at 4:04
  • $\begingroup$ @Benny which you know are the same. $\endgroup$ – Igor Rivin Mar 6 '18 at 4:04
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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.

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    $\begingroup$ I may have thought in the very same way, if the equation on page 25 wasn’t given so directly. Thanks, you got what I exactly want to know! $\endgroup$ – Hello Mar 6 '18 at 4:30

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