# Commutator subgroup

Commutator subgroup

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.

## 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.

## Definition

This motivates the definition of the commutator subgroup [G,G] (also called the derived subgroup, and denoted G′ or G(1)) of G: it is the subgroup generated by all the commutators.

It follows from the properties of commutators that any element of [G,G] is of the form

$[g_1,h_1] \cdots [g_n,h_n]$

for some natural number n. Moreover, since

$([g_1,h_1] \cdots [g_n,h_n])^s = [g_1^s,h_1^s] \cdots [g_n^s,h_n^s]$, the commutator subgroup is normal in G. For any homomorphism $f: G \rightarrow H$,

$f([g_1,h_1] \cdots [g_n,h_n]) = [f(g_1),f(h_1)] \cdots [f(g_n),f(h_n)]$,

so that $f([G,G]) \leq [H,H]$.

This shows that the commutator subgroup can be viewed as a functor on the category of groups, some implications of which are explored below. Moreover, taking G = H it shows that the commutator subgroup is stable under every endomorphism of G: that is, [G,G] is a fully characteristic subgroup of G, a property which is considerably stronger than normality.

The commutator subgroup can also be defined as the set of elements g of the group which have an expression as a product g = g1g2...gk that can be rearranged to give the identity.

### Derived series

This construction can be iterated:

G(0): = G
$G^{(n)} := [G^{(n-1)},G^{(n-1)}] \quad n \in \mathbb{N}$

The groups $G^{(2)}, G^{(3)}, \ldots$ are called the second derived subgroup, third derived subgroup, and so forth, and the descending normal series

$\cdots \triangleleft G^{(2)} \triangleleft G^{(1)} \triangleleft G^{(0)} = G$

is called the derived series. This should not be confused with the lower central series, whose terms are Gn: = [Gn − 1,G], not G(n): = [G(n − 1),G(n − 1)].

For a finite group, the derived series terminates in a perfect group, which may or may not be trivial. For an infinite group, the derived series need not terminate at a finite stage, and one can continue it to infinite ordinal numbers via transfinite recursion, thereby obtaining the transfinite derived series, which eventually terminates at the perfect core of the group.

### Abelianization

Given a group G, a factor group G/N is abelian if and only if [G,G] ≤ N.

The quotient G / [G,G] is an abelian group called the abelianization of G or G made abelian. It is usually denoted by Gab or Gab.

There is a useful categorical interpretation of the map $\phi: G \rightarrow G^{\operatorname{ab}}$. Namely ϕ is universal for homomorphisms from G to an abelian group H: for any abelian group H and homomorphism of groups $f: G \rightarrow H$ there exists a unique homomorphism $F: G^{\operatorname{ab}} \rightarrow H$ such that $f = F \circ \phi$. As usual for objects defined by universal mapping properties, this shows the uniqueness of the abelianization $G^{\operatorname{ab}}$ up to canonical isomorphism, whereas the explicit construction $G \rightarrow G/[G,G]$ shows existence.

The abelianization functor is the left adjoint of the inclusion functor from the category of abelian groups to the category of groups.

Another important interpretation of $G^{\operatorname{ab}}$ is as $H_1(G,\mathbb{Z})$, the first homology group of G with integral coefficients.

### Classes of groups

A group G is an abelian group if and only if the derived group is trivial: [G,G] = {e}. Equivalently, if and only if the group equals its abelianization. See above for the definition of a group's abelianization.

A group G is a perfect group if and only if the derived group equals the group itself: [G,G] = G. Equivalently, if and only if the abelianization of the group is trivial. This is "opposite" to abelian.

A group with G(n) = {e} for some n in N is called a solvable group; this is weaker than abelian, which is the case n = 1.

A group with G(α) = {e} for some ordinal number, possibly infinite, is called a hypoabelian group; this is weaker than solvable, which is the case α is finite (a natural number).

## Examples

### Map from Out

Since the derived subgroup is characteristic, any automorphism of G induces an automorphism of the abelianization. Since the abelianization is abelian, inner automorphisms act trivially, hence this yields a map

$\mbox{Out}(G) \to \mbox{Aut}(G^{\mbox{ab}})$

## References

1. ^ Dummit, David S.; Foote, Richard M. (2004). Abstract Algebra (3rd ed.). John Wiley & Sons. ISBN 0-471-43334-9.
2. ^ Lang, Serge (2002). Algebra. Graduate Texts in Mathematics. Springer. ISBN 0-387-95385-X.

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