## Review of Group Theory: The Commutator Subgroup and the Abelianization of a Group

**Point of post: **In this post we discuss the notion of the commutator subgroup of a group and prove some basic, yet important, facts about it. We then discuss the abelianization of a group (the group mod the commutator subgroup).

*Motivation*

Very often in group theory we are dealt groups which, to our consternation, are not abelian. That said, being abelian and not abelian is often too black-and-white of a measure of commutativity. Thus, one may ask if there is some in-between quality that measures the ‘abelianess’ or lack thereof of the group. There is–this is the commutator subgroup. In essence the commutator subgroup is precisely such that when the ambient group is quotiented out by it everything of the form goes to the identity. This makes sense since in a perfect world, an abelian world that is, everything of this form would already be the identity. Thus, quotienting out by the subgroup generated by the set of all (i.e. the commutator subgroup) collapses all the defiant breakers of the golden abelian rule. Thus, in essence the bigger the commutator subgroup is the further from abelian the group is.

*Commutator Subgroup*

** **Let be any group. We define the map given by to be the

*commutator bracket.*Then the subgroup generated by the set is called the

*commutator subgroup of*and is denoted . Our first claim is that is a normal subgroup of . Indeed it clearly suffices to check that for every . Recall the inner automorphism and the fact that it is, surprisingly, an automorphism. Thus, one sees that and since were arbitrary it follows that as claimed.

*Abelinization*

Now that we know we may speak freely of quotient groups. We call the quotient group the *abelinization *of and denote it . The first interesting theorem pertaining to is:

**Theorem: ***Let be any group and . Then, is abelian if and only if .*

**Proof: **Suppose first that and let be the canonical projection. We note then that for any one has that (since ) and so . Then, since is surjective the conclusion follows.

Conversely, suppose that is abelian. Then, for any one has that or

which is true if and only if . Since were arbitrary it follows that as desired.

**Corollary: *** is abelian.*

The really interesting theorem regarding though is:

**Theorem: ***Let and be any two groups. Then a homomorphism has the property that is abelian if and only if can be factored through by , in other words if and only if there exists a homomorphism such that .*

**Proof: **Clearly if can be factored in such a way then and is the image of an abelian group under a homomorphism, and thus trivially abelian.

Conversely, suppose that is abelian. Consider then the map by . This is well-defined since if then for some . But, since is abelian it’s fairly easy to see that and so . Furthermore, it’s evident that is a homomorphism and .

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2009

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