Derived group
From Exampleproblems
In mathematics, the derived group (or commutator subgroup) of a group G is the subgroup G1 generated by all the commutators of elements of G; that is, G1 = <[g,h] : g,h in G>.
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.
Note that the set of all commutators of the group is, generally, not a group (in any interesting case). While clumsily defined, the commutator subgroup is important.
An abelian group has only trivial commutators. Hence its commutator subgroup is {1}. The converse is also (trivially) true.
The derived group, in a sense, gives a measure of how far G is from being abelian; the larger G1, the "less abelian" G is. In particular, G1 is equal to {1} if and only if the group G is abelian. A perfect group G is one with G1 = G.
If f : G -> H is a group homomorphism, then f(G1) is a subgroup of H1, because f maps commutators to commutators. This implies that the operation of forming derived groups is a functor from the category of groups to the category of groups.
Applying this to endomorphisms f, we find that G1 is a fully characteristic subgroup of G, and in particular a normal subgroup of G. The quotient G/G1 is an abelian group sometimes called G made abelian, or the abelianization of G. In a sense, it is the abelian group that's "closest" to G, which can be expressed by the following universal property: if p : G -> G/G1 is the canonical projection, and f : G -> A is any homomorphism from G to an abelian group A, then there exists exactly one homomorphism s : G/G1 -> A such that s o p = f. In the language of category theory: the functor which assigns to every group its abelianization is left adjoint to the forgetful functor which assigns to every abelian group its underlying group.
In particular, a quotient G/N of G is abelian if and only if N includes G1.de:Kommutatorgruppe fr:Groupe dérivé ru:Коммутант
