Abstract Nonsense

Crushing one theorem at a time

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 G 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 xyx^{-1}y^{-1} 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 xyx^{-1}y^{-1} (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.

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Commutator Subgroup

Let G be any group. We define the map [\cdot,\cdot]:G\times G\to G given by [x,y]=xyx^{-1}y^{-1} to be the commutator bracket. Then the subgroup generated by the set \left\{[x,y]:x,y\in G\right\} is called the commutator subgroup of G and is denoted \left[G,G\right]. Our first claim is that [G,G] is a normal subgroup of G. Indeed it clearly suffices to check that g[x,y]g^{-1}\in [G,G] for every g,x,y\in G. Recall the inner automorphism i_g and the fact that it is, surprisingly, an automorphism. Thus, one sees that gxyx^{-1}y^{-1}g^{-1}=i_g(xyx^{-1}y^{-1})=i_g(x)i_y(y)i_g(x)^{-1}i_g(y)^{-1}\in [G,G] and since g,x,y\in G were arbitrary it follows that [G,G]\unlhd G as claimed.

Abelinization

Now that we know [G,G]\unlhd G we may speak freely of quotient groups. We call the quotient group G/[G,G] the abelinization of G and denote it G^{\text{ab}}. The first interesting theorem pertaining to G^{\text{ab}} is:

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Theorem: Let G be any group and N\unlhd G. Then, G/N is abelian if and only if [G,G]\leqslant N.

Proof: Suppose first that [G,G]\leqslant N and let \pi:G\to G/N be the canonical projection. We note then that for any x,y\in G one has that \pi(x)\pi(y)\pi(x)^{-1}\pi(y)^{-1}=\pi\left(xyx^{-1}y^{-1}\right)=N (since xyx^{-1}y^{-1}\in[G,G]\subseteq N) and so \pi(x)\pi(y)=\pi(y)\pi(x). Then, since \pi is surjective the conclusion follows.

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Conversely, suppose that G/N is abelian. Then, for any x,y\in G one has that \pi(x)\pi(y)=\pi(y)\pi(x) or
\pi\left(xyx^{-1}y^{-1}\right)=N which is true if and only if xyx^{-1}y^{-1}\in N. Since x,y\in G were arbitrary it follows that [G,G]\subseteq N as desired. \blacksquare

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Corollary: G^{\text{ab}} is abelian.

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The really interesting theorem regarding G^{\text{ab}} though is:

Theorem: Let G and H be any two groups. Then a homomorphism \phi:G\to H has the property that \text{im}(\phi)\leqslant H is abelian if and only if \phi can be factored through \text{G}^{\text{ab}} by \pi, in other words if and only if there exists a homomorphism \theta:\text{G}^{\text{ab}}\to H such that \phi=\theta\circ\pi.

Proof: Clearly if \phi can be factored in such a way then \text{im}(\phi)=\text{im}(\theta) and \text{im}(\theta) is the image of an abelian group under a homomorphism, and thus trivially abelian.

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Conversely, suppose that \text{im}(\phi) is abelian. Consider then the map \displaystyle \theta:G^{\text{ab}}\to H by g[G,G]\mapsto \phi(g). This is well-defined since if g[G,G]=g'[G,G] then g'=gk for some k\in [G,G]. But, since \text{im}(\phi) is abelian it’s fairly easy to see that \phi(k)=e and so \phi(g')=\phi(g). Furthermore, it’s evident that \theta is a homomorphism and \phi=\theta\circ\pi. \blacksquare

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References:

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

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February 27, 2011 - Posted by | Algebra, Group Theory | , , , , ,

8 Comments »

  1. […] Recall from our discussion of the abelinization of a group that for every group and every homomomorphism such that is […]

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  5. […] We recall that the dual group  has order equal the the abelianization . But, by inspection we see that and so and so (the commutator subgroup). It evidently follows […]

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