The Derived Subgroup

Relevant Classes: Math 493

Definition: The Derived Subgroup

Let $G$ be a group. The derived subgroup $[G,G]$ of $G$ is the subgroup generated by its commutators: $[G,G]=⟨xy{x}^{-1}{y}^{-1}\mid x,y\in G⟩.$

This subgroup has a few interesting and useful properties.

Normality

Fix an element in $[G,G]$. It is of the form $xy{x}^{-1}{y}^{-1}$. Thus, we have:

\begin{align*} (xyx^{-1}y^{-1})g^{-1} &= gxg^{-1} gyg^{-1} gx^{-1}g^{-1} gy^{-1}g^{-1} \\ &= (gxg^{-1}) (gyg^{-1}) (gxg^{-1})^{-1} (gyg^{-1})^{-1} \end{align*}

which is another commutator, and thus is in $[G,G]$.

Since $[G,G]◃G$, we can take the quotient group $G/[G,G]$. This has a few interesting properties:

1. $G/[G,G]$ is abelian.
2. If $G/N$ is abelian, then $[G,G]\subseteq N$. Thus, $G/[G,G]$ is the largest abelian quotient group.

Abelian

Fix two cosets, $x[G,G]$ and $y[G,G]$. We want to show that $(x[G,G])(y[G,G])=(y[G,G])(x[G,G]).$

(x[G, G]) (y[G, G]) &= x y [G, G] \\ &= xy (y^{-1} x^{-1} y x [G, G]) \\ &= y x [G, G] \\ &= (y[G, G]) (x[G, G]). \end{align*} $$ Thus, $G/[G, G]$ is abelian.

Largest Abelian Subgroup

Suppose $N$ is a normal subgroup of $G$ and $G/N$ is abelian. Then, we have $(xN)(yN)=(yN)(xN)$ which means that $xyN=yxN$, which gives us that $y^{-1}{x}^{-1}yxN=N⟹y^{-1}{x}^{-1}yx\in N.$

That means that the derived subgroup $[G,G]\le N$.