In $G/K(G)$ for commutator group $K(G)$, it is said that $[a][b]=[a][b][b^{-1}a^{-1}ba]=[a][b][b^{-1}][a][b][a]=[b][a].$ Why the first equality?

Question

Let $K(G)$ be the commutator subgroup. It is said that in the quotient space $G/K(G)$ $$\begin{align} [a][b]&=[a][b][b^{-1}a^{-1}ba]\\ &=[a][b][b^{-1}][a][b][a]\\ &=[b][a]. \end{align}$$

Where does the first equality come from?

I only see that since $ab=[a,b]ba$, I get $$\begin{align} [a][b]&=[ab]\\ &=[[a,b]ba]\\ &=[aba^{-1}b^{-1}ba]\\ &=[a][b][a^{-1}b^{-1}ba] \end{align}$$

In the abelianization $b^{-1}a^{-1}ba=e,$ because you mod out by the commutator. Not in the commutator itself. It's often not abelian. — suckling pig, Jan 21 '23 at 16:50
Ok, by $K(G)$ do you mean the commutator? Because you have a quotient there. — suckling pig, Jan 21 '23 at 16:56
No, by K(G) I mean the commutator subgroup, i.e. the subgroup generated by all the $[a,b]$ with $a,b\in G$. — BridgeTYH, Jan 21 '23 at 16:58
That's what I meant. $K(G)$, the commutator subgroup, is not always abelian. If you take the quotient, $G/K(G)$, which is what you wrote, you get an abelian group. — suckling pig, Jan 21 '23 at 17:00
Yes, but I want to show that it is abelian. So why do I have $b^{-1}a^{-1}ba=e$? — BridgeTYH, Jan 21 '23 at 17:01
Look up "quotient group". For instance, $\Bbb Z/n\Bbb Z.$ How does it work? — suckling pig, Jan 21 '23 at 17:03
It's not $b^{-1}a^{-1}ba$ that's $e$; rather, $[b^{-1}a^{-1}ba]=[e]$. — Shaun, Jan 21 '23 at 17:09

suckling pig · Accepted Answer · 2023-01-21T18:06:12.893

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You can't prove that the commutator subgroup, your $K(G),$ is abelian, because in general it isn't.

For instance, $K(S_n)=A_n$, which is not abelian ($n\gt3$).

What you have above is the proof that $G/K(G)$ is abelian. That's the quotient of $G$ by $K(G)$. It's called the abelianization.

Since it's the first equality that bothers you, just note that in this quotient, everything in the commutator subgroup is set equal to $\bar e.$ But the commutator subgroup contains each $b^{-1}a^{-1}ba.$

edited Jan 21 '23 at 18:06

answered Jan 21 '23 at 17:18

suckling pig

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I don't think this answers the question. – Shaun Jan 21 '23 at 17:36
@Shaun It does answer the question. the square brackets in the OP mean "equivalence class mod commutators'. so, the first equality is multiplying by the identity. – Igor Rivin Jan 21 '23 at 18:19
Well, it does now, @IgorRivin; it didn't before. – Shaun Jan 21 '23 at 18:30

In $G/K(G)$ for commutator group $K(G)$, it is said that $[a][b]=[a][b][b^{-1}a^{-1}ba]=[a][b][b^{-1}][a][b][a]=[b][a].$ Why the first equality?

1 Answers1