# Commutator subgroup

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In mathematics, more specifically in abstract algebra, the commutator subgroup or derived subgroup of a group is the subgroup generated by all the commutators of the group.[1][2]

The commutator subgroup is important because it is the smallest normal subgroup such that the quotient group of the original group by this subgroup is abelian. In other words, G/N is abelian if and only if N contains the commutator subgroup. So in some sense it provides a measure of how far the group is from being abelian; the larger the commutator subgroup is, the "less abelian" the group is.

## Contents

### Commutators

For elements g and h of a group G, the commutator of g and h is [g,h]: = g − 1h − 1gh. The commutator [g,h] is equal to the identity element e if and only if gh = hg, that is, if and only if g and h commute. In general, gh = hg[g,h].

An element of G which is of the form [g,h] for some g and h is called a commutator. The identity element e = [e,e] is always a commutator, and it is the only commutator if and only if G is abelian.

Here are some simple but useful commutator identities, true for any elements s, g, h of a group G:

• [g,h] − 1 = [h,g].
• [g,h]s = [gs,hs], where gs = s − 1gs.

For any homomorphism $f: G \rightarrow H$, f([g,h]) = [f(g),f(h)].

The first and second identities imply that the set of commutators in G is closed under inversion and under conjugation. If in the third identity we take H = G, we get that the set of commutators is stable under any endomorphism of G. This is in fact a generalization of the second identity, since we can take f to be the conjugation automorphism $x \mapsto x^s$.

However, the product of two or more commutators need not be a commutator. A generic example is [a,b][c,d] in the free group on a,b,c,d. It is known that the least order of a finite group for which there exists two commutators whose product is not a commutator is 96; in fact there are two nonisomorphic groups of order 96 with this property.