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Let $(G,\cdot)$ be a group and define the comutator as $[a,b] := aba^{-1}b^{-1}$ for every $a,b\in G$. About to prove that $[G,G] \leqslant_n G$, first i need to prove that is in fact a subgroup, wich is were i failed. My question is:

Given $a,b,c,d\in G$, find $x,y\in G$ such $[a,b]\cdot[c,d] = [x,y]$.

This is for proving that $[\,\cdot\,,\,\cdot\,]$ are closed at $[G,G]$. The otter statments wich defines a group i've allready check it, but they doesn't defines a group if the operation $\cdot$ wouldn't give me an element of $[G,G]$. Thanks for the help.

Mistakes. In fact, $[G,G]$ isn't the comutator as well, so $[G,G] {\require{cancel}\cancel{\leqslant}} G$, so, the product of the comutators isn't a comutator as well

Gustavo Mezzovilla
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    This is not the same question at all. The question in the link is about proving that it is a normal subgroup. The question here is asking help in proving something that is not true. – Derek Holt Oct 09 '19 at 14:47
  • I think you have misunderstood exactly what the commutator subgroup is. The product of two commutators is not necessarily a commutator. – Arthur Oct 09 '19 at 14:49
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    I think there is a misunderstanding here: the product of two commutators may not be a commutator. However, the commutator subgroup is not the collection of all commutators, it is the collection of all finite products of commutators; that is, the commutator subgroup is the subgroup generated by the commutators. – User8128 Oct 09 '19 at 14:49
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    Wow, so $[G,G] \neq \big{[a,b] \mid a,b\in G\big}$ ? – Gustavo Mezzovilla Oct 09 '19 at 14:51
  • Can we define $[G,G]$ to be the smallest subgroup $H\leqslant G$ such that for every $x\in H$, exists and $n\in\Bbb N$ and $\Omega \subset {[a,b]\mid a,b\in G}$ such that $|\Omega| = n$ and $x = \prod\limits_{g\in \Omega} g$ ? – Gustavo Mezzovilla Oct 09 '19 at 14:56
  • If I've read your comment correctly, your group $H$ will always be the trivial group – Robert Chamberlain Oct 09 '19 at 15:46
  • In general, $[G,G]$ need not be equal to the set of commutators; though the smallest examples are large-ish. There are several papers exploring just how large a group has to be before the set of commutators is not the commutator subgroup. Your "definition" doesn't work, but you can ask, for a given $G$, what is the smallest $n$ for which every element of $[G,G]$ is a product of at most $n$ commutators. For example, the Ore Conjecture (now a theorem) was that for a finite nonabelian simple group, $n=1$ (that is, every element is in fact a commutator). – Arturo Magidin Oct 09 '19 at 16:56

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This is a common error.

In a group $G$, with $a,b\in G$, the commutator of $a$ and $b$ is $[a,b]=aba^{-1}b^{-1}$ (note: it can also be defined as $[a,b]=a^{-1}b^{-1}ab$; some formulas change a bit, but the basic properties are the same; the difference is that with your definition, $[a,b]$ is the unique element of $G$ such that $ab=[a,b]ba$, whereas with the other definition it is the unique element such that $ab=ba[a,b]$).

However, the commutator subgroup $[G,G]$ of $G$ is not the set of all commutators. It is the subgroup generated by all commutators. That is, $$[G,G] = \Bigl\langle [a,b]\,\Bigm|\,a,b\in G\Bigr\rangle,$$ and not $\{[a,b]\mid a,b\in G\}$.

In fact, it is false that the set of commutators is a subgroup in general, though specific examples may be difficult to find, as the smallest group for which the set of commutators is not a subgroup is of order 96; this was established in Robert Guralnick's dissertation. In fact, there are two nonisomorphic groups of order 96 in which the set of commutators is not closed under products.

There is a nice note of Marty Isaacs in the Monthly providing an infinite family of examples of finite groups in which the commutator subgroup contains elements that are not commutators: you can find it here courtesy of Pete L. Clark. It is

  • Commutators and the commutator subgroup by I.M. Isaacs, American Mathematical Monthly 84 (no. 9) (Nov 1977), pp. 720-722

More information in this math.overflow question and this question on this very site

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Arturo Magidin
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