The commutator of two group elements *a* and *b* is the group element *c* such
that *ab* = *bac*. It can be thought of as that element which allows *a* and
*b* to commute. It may also be defined as
*Comm*(*a*,*b*) = *a*^{-1}*b*^{-1}*ab*; it
should be easy to see that this satisfies the role of *c*. It should also be
clear from both definitions that if *a* and *b* commute then *c* is *I*, and
*Comm*(*a*,*b*) will come out to be *I*. If the group is abelian so that all
pairs *a*,*b* commute then the only commutator is *I* and the commutator
subgroup is *I*. The reverse is also true.

Commutators are preserved by homomorphisms.

and remembering that homomorphisms map inverses to inverses we find

which is a commutator. So if the commutators and their products form a proper subgroup of the group, all endomorphisms will preserve this subgroup, and so it is fully invariant, characteristic, and normal.

The factor group of commutator subgroups is always abelian. It may help to
recall what the factor group is, namely where a normal subgroup is mapped to
the identity of some group and its cosets are mapped to the other elements of
that group. So multiplication in the factor group can be thought of as
membership in cosets of the starting group. If *N* is a normal subgroup of
*G*, *aN* is in one coset and *bN* is in another, then in general *aNbN* goes
to one coset of *G*, and *bNaN* may go to yet another. But not when *N*=*C*.

For multiplication of cosets to be commutative we need *ab* and *ba* to go to
the same coset, *abC* = *baC*, or in terms of concrete elements
*abc*_{1} = *bac*_{2}.
But we can take *c*_{1} to be *I* and get *ab* = *bac*, which is exactly the
definition of commutators! If that was too fast there is a longer derivation:
*aCbC* = *abCC* = *abC* = *baCC* = *bCaC*.

A third albeit currently less rigorous way of looking at it is that in making
the factor group we have gathered up all the commutators of *G* into *C* and
then mapped them all to *I*. For the factor group to not be abelian one of
the other elements of *G*/*C* must be a commutator (if the only commutator is
*I* the group is abelian). But the pre-image of this commutator cannot be a
commutator or product of commutators in *G*, because those have all be mapped
to *I*. Since homomorphism preserves group structure one might suspect this
might not be the case. In a sense we have gathered all the non-commutativity
of *G* into a bag and then squished the bag.