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I know that a commutator subgroup $C$ of a group $G$ is defined as $C=\{aba^{-1}b^{-1}|a,b\in G\}$. When I reading the textbook, it always state that $C$ is a subgroup of $G$ and only prove if it is normal.

To prove a set is a group, we need to prove is it closure, associative, have an identity element and has inverse, but when I check my textbook on how to prove $C$ is a group, it only shows identity element and inverse, although associative is too obvious, but as the closure I cannot find a way to prove that.

The question is, how to prove that $aba^{-1}b^{-1}hkh^{-1}k^{-1}\in C$ for all $a,b,h,k\in G$?

kelvin hong 方
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    That's not the definition of the commutator subgroup. The commutator subgroup is defined as the subgroup generated by the commutators. – Angina Seng Aug 26 '18 at 06:50
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    i.e. $C = \langle aba^{-1}b^{-1} \mid a,b\in G\rangle$ – Andre of Astora Aug 26 '18 at 06:53
  • @LordSharktheUnknown You mean there are some elements in $C$ cannot express in the form $aba^{-1}b^{-1}$ right? It is just 'generated' by it but not all of them have the form? – kelvin hong 方 Aug 26 '18 at 06:57
  • @postmortes Sir can you expressed explicitly how $aba^{-1}cb^{-1}c^{-1}$ can be the form $xyx^{-1}y^{-1}$? Additionally, it seems that the two elements you choose have common variable $b$, can you write how can I write the product of the two with no common variable? – kelvin hong 方 Aug 26 '18 at 07:21
  • @postmortes This sounds stranges, can you show me what other things I should have, to prove I can actually do that? – kelvin hong 方 Aug 26 '18 at 07:33
  • Actually, I've just found a simple example that shows I was wrong in my first claim, so I'm deeply sorry about that. Consider the group $SL_2({\mathbb R})$. $-I$ (where I is the $2\times 2$ identity matrix) belongs to the commutator subgroup but cannot be expressed as a commutator. – postmortes Aug 26 '18 at 07:38
  • @postmortes Its okay... Is $I$ here a product of commutators? – kelvin hong 方 Aug 26 '18 at 07:45
  • No. That's the point. $-I$ belongs to the commutator subgroup (because the commutator subgroup of $SL_2({\mathbb R})$ is actually $SL_2({\mathbb R})$) but cannot be expressed as the product of commutators. – postmortes Aug 26 '18 at 07:51
  • @postmortes How about consider $I=III^{-1}I^{-1}$? It seems like a commutator. – kelvin hong 方 Aug 26 '18 at 07:54
  • @kelvinhong方 But $I$ is not $-I$, so we're not talking about the same thing. – postmortes Aug 26 '18 at 12:04
  • Let us continue this discussion in chat. – kelvin hong 方 Aug 26 '18 at 12:08

1 Answers1

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In general, the set of commutators does not equal the commutator subgroup, as pointed out above by several remarks. See also this paper of Kappe and Morse, where criteria are discussed for equality.

I. D. MacDonald provides counterexamples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]. In particular, he proves by a simple counting argument the following theorem.

If $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.

Finally see also this StackExchange entry, which mentions another interesting paper of Marty Isaacs (also behind Jstor) with more counterexamples.

Nicky Hekster
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