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I've read wikipedia, tried examples ($S_3$ stuff) but I cannot understand why they are called "commutators" or what the set of them is.

Why is the set of commutators called "commutator subgroup"? (I read this one among other, I have done reading even I don't link all the tabs!)

It's in Lang's Undergraduate Algebra - page 35, question 10 part b. It's in A Book of Abstract Algebra by Pinter (page 152), but I still have no idea what this is.

My goal is to show that if $H$ is a subgroup of $G$ that contains all the commutators that $G/H$ is Abelian - please don't do this for me. I'm stuck with trying to work out what H is not the question.

See... if the result is true ($G/H$ being Abelian) then H is the set of things that don't commute because I'm factoring them out with the quotient operation. The identity of H will be left in the result (necessarily) thus the commutators that = the identity (ie the things that actually commute) are left in the group.

So is this set of commutators $=\{x|x=aba^{-1}b^{-1}, \forall a,b\in G\}$ the reason I need this clarified is because then an Abelian group only has one commutator, the identity.

Also the commutators don't actually commute! For example $(1,3)$ and $(1,3)(1,2)$ in $S_3$. Have I actually got the definition and the name has just confused me? Also both books say "a commutator of G is an element of the form" but there might only be one (for an Abelian one)

Could someone clarify please?

Oh group generated means "the union of all the groups generated by each element of the set, generated as in $<g>=\{e,g,g^2....\}$"

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    $\begingroup$ I think of it as the 'group element that prevents two elements from commuting', since $$ab = ba[a,b]$$If that darned thing weren't there, a and b would commute. $\endgroup$ – David Peterson Sep 5 '14 at 21:55
  • $\begingroup$ @DavidPeterson that's brilliant, one quick thing, I've spotted the set of commutators is not a group, but the group generated by... how do I define that? It's the union of $<[a,b]>$ right? for all the different $[a,b]$s $\endgroup$ – Alec Teal Sep 5 '14 at 21:58
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    $\begingroup$ @AlecTeal Do you know the definition of "subgroup generated by a set $S$"? It is the set of all finite products of elements of your set $S$, and their inverses. Since $[a,b]^{-1}=[b,a]$, the subgroup generated by commutators is the set of all finite products of commutators, $[G,G]=\{p:p\text{ is a finite product of commutators}\}$. Can you prove $[G,G]$ is a subgroup? Can your prove $G/[G,G]$ is abelian? Hint: $ab[G,G]=ba[G,G]$ iff $b^{-1}a^{-1}ba[G,G]=[G,G]$ iff $b^{-1}a^{-1}ba\in [G,G]$ for every $a,b$. Note that $a[b,c]a^{-1}=[aba^{-1},aca^{-1}]$, so $[G,G]$ is normal. $\endgroup$ – Pedro Tamaroff Sep 5 '14 at 22:01
  • $\begingroup$ Note that the set of commutators is not, in general, a subgroup. For instance, consider the free group on two generators. The commutator subgroup is the one generated by the set of commutators. $\endgroup$ – Moishe Kohan Sep 5 '14 at 22:04
  • $\begingroup$ @PedroTamaroff quick question, a commutator raised to a power is just the commutator! Anyway, the "group generated by" is just $\{x^n|x\in S, n\in\mathbb{Z}\}$ right? $\endgroup$ – Alec Teal Sep 5 '14 at 22:29
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If $a,b\in G$ then $ab=ba$ is equivalent to $aba^{-1}b^{-1}=e$ and so the commutator of $a$ and $b$ is a measure of the failure of $a$ and $b$ commuting with each other. However, as a measure alone, it is somewhat binary: either group elements commute or they don't.

There are two ways we can quantify this measure. The first is by looking at (subgroups of) the symmetric group. Then, instead of asking if $a$ and $b$ commute, we can ask how many elements are moved around by their commutator.

The second way is to look at the commutator subgroup as a measure of how noncommutative a group is. A group is commutative if it has a trivial commutator subgroup (and highly noncommutative if the commutator subgroup is the entire group). But we can then ask if the commutator subgroup is abelian. And if not, if it's commutator subgroup is abelian. And we can repeat the process of taking commutator subgroups many times and ask if we ever find the trivial group, and if so, how many times does it take? And if not, how many times does it take before we keep on getting the same group repeating itself? This gives the notion of solvable groups and their degree of sol ability. A similar notion gives nilpotent groups and their degree of nilpotency. Both are measures of how close to abelian our groups actually are.

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    $\begingroup$ More generally (from your second paragraph) if you have a compatible metric on the group (like $d(a,b)$ equal to size of support of $ab^{-1}$ in a permutation group, as in your example), one can measure the distance of the commutator to $e$. $\endgroup$ – tomasz Sep 5 '14 at 22:18
  • $\begingroup$ In fact, if we take a discrete metric (which is certainly compatible with any group structure), then we get this binary thing you mention in first paragraph, so that is a special case as well. $\endgroup$ – tomasz Sep 5 '14 at 22:34
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We call it commutator, because it shows us the operation in the group is "how far" from commutativity.

Let $(G,\cdot)$ be a group. For all $g,h \in G$ we have $[g,h]=g^{-1}h^{-1}gh=e$ if and only if the group $(G,\cdot)$ is commutative.

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I don't like this talk of "measuring the failure" of commutativity. (Admittedly, that is just my preference, but I don't see any natural notion of measurement here.)

I like to think of group elements by their actions, and a commutator $c$ for elements $a$ and $b$ is a thing $c$:

$$ ab = bac $$

That is, $c$ is the group action that makes $b$ and $a$ commute. Of course, if $a$ and $b$ commute naturally, either because the group is commutative, or they just happen to commute, then $ab = ba$ and $c = e$.

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