Commutator Subgroup

Subjects: Group Theory
Links: Normal Subgroups and Quotient Groups, Subgroups, Cauchy and Sylow Theorems

Def: Let $G$ be a group, and $a,b\in G$, then the commutator of $a$ and $b$ is $[a,b]:=a^{-1}{b}^{-1}ab\in G$. The commutator is the element $e$ iff $ab=ba$, that is $a$ and $b$ commute. In general, $ab=ba[a,b]$, like how close are those elements to commute. An element of $G$ of the form $[a,b]$ is called a commutator.

The commutator subgroup $[G,G]$ or the derived subgroup denoted as $G^{\prime }$ or $G^{(1)}$: it is the subgroup generated by all the commutators. By definition, any element of $[G,G]$ is of the form $$[a_1, b_1] \cdots [a_n, b_n]$$for some $n\in \mathbb{N}$, and where $a_i,b_i\in G$. Doing some elementary algebra we get that $[G,G]⊴G$.

Prop: Let $G$ be a group and $H⊴G$. $G/H$ is abelian iff $[G,G]\le H$.

In particular $G/[G,G]$ is abelian and it is called the abelianisation of $G$, usually denoted as $G^{\text{ab}}$ or $G_{\text{ab}}$.

A group $G$ is called perfect if $G=[G,G]$. This like the opposite of being abelian.

Characteristic Property of the Abelianisation: Let $G$ be a group. For any abelian group $H$ and any homomorphism $\phi :G\to H$, there exists a unique homomorphism $g:G^{\text{ab}}\to H$ such that the following diagram commutes:

\usepackage{tikz-cd}
\usepackage{amsfonts, amsmath, amssymb}
\begin{document} 
\begin{tikzcd}[row sep=2cm, column sep=2cm]
G \arrow[d, two heads,"\varphi"'] \arrow[r, "f"] & H \\
{G^\text{ab}}\arrow[ur, dashed, "g"']
\end{tikzcd}
\end{document}

Cor: Given a group $G$, then $G^{\text{ab}}$ is the unique group (up to isomorphism) that satisfies the characteristic property.

Cor: We see that the abelianisation defines a covariant functor from $\mathsf{G}\mathsf{r}\mathsf{p}$ to $\mathsf{A}\mathsf{b}$. This is because if $f:G\to H$ is a group homomorphism, then there exists a unique homomorphism $f^{\text{ab}}:G^{\text{ab}}\to H^{\text{ab}}$ such that the following diagram commmute:

\usepackage{tikz-cd}
\usepackage{amsfonts, amsmath, amssymb}
\begin{document} 
\begin{tikzcd}[row sep=2cm, column sep=2cm]
G \arrow[d, two heads,"\pi"'] \arrow[r, "f"]\arrow[dr, "\pi \circ f"] & H \arrow[d, two heads,"\pi"] \\
{G^\text{ab}}\arrow[r, dashed, "f^\text{ab}"'] & H^\text{ab}
\end{tikzcd}
\end{document}

Thus, $(-{)}^{\text{ab}}:\mathsf{G}\mathsf{r}\mathsf{p}\to \mathsf{A}\mathsf{b}$ is a covariant functor, where $G\mapsto G^{\text{ab}}$, and $(f:G\to H)\mapsto f^{\text{ab}}:G^{\text{ab}}\to H^{\text{ab}}$.

Prop: Let $\{G_{\alpha }\mid \alpha <\kappa \}$ be a collection of groups. Then $$\left(\coprod_{\alpha<\kappa} G_\alpha\right)^\text{ab} \cong \bigoplus_{\alpha <\kappa} G_\alpha^\text{ab}.$$This means that $(-{)}^{\text{ab}}$ sends the coproduct of $\mathsf{G}\mathsf{r}\mathsf{p}$ to the coproduct of the images in $\mathsf{A}\mathsf{b}$.

Cor: Let $S$ be a nonempty set, then $F(S{)}^{\text{ab}}\cong \mathbb{Z}⟨S⟩$, where $F(S)$ is the free group generated by $S$, and $⟨S⟩$ is the free abelian group generated by $S$, or the free $\mathbb{Z}$-module generated by $S$.

Prop: Let $\{G_{\alpha }\mid \alpha <\kappa \}$ be a collection of groups. Then $$\left(\prod_{\alpha<\kappa} G_\alpha\right)^\text{ab} \cong \prod_{\alpha <\kappa} G_\alpha^\text{ab}.$$This means that $(-{)}^{\text{ab}}$ sends the product of $\mathsf{G}\mathsf{r}\mathsf{p}$ to the product of the images in $\mathsf{A}\mathsf{b}$.

Def: Let $G$ be a group. We define the following groups inductively $G^{(0)}:=G$, and $$G^{(n+1)} := [G^{(n)}, G^{(n)}]$$for $n\ge 1$. The groups $G^{(2)}$, $G^{(3)},\dots$ are called the second derived subgroup, third derived subgroup and so forth.

We get the following properties for the derived subgroups:

• $G^{(n+1)}⊴G^{(n)}$ for every $n\ge 0$.
• $G^{(n+1)}⊴G$ for every $n\ge 0$.
• The quotients $G^{(n)}/G^{(n+1)}$ is abelian for every $n\ge 0$.

Prop: A group $G$ is solvable iff $G^{(n)}=1$ for some $n\ge 0$.

Th: Let $G$ be a group. Then,

• If $H\le G$ is a subgroup and $G$ is solvable, then $H$ is solvable.
• If $H⊴G$ and $G$ is solvable, then $G/H$ is solvable.
• If $H⊴G$ is such that $H$ and $G/H$ are solvable, then $G$ is solvable.

This means that the class $\mathsf{S}\mathsf{o}\mathsf{l}$ of solvable groups is closed under subgroups, quotients and extensions.

We can extend the definition of the commutator subgroup. If $H$ and $K$ are two subgroups of $G$, we denote $[H,K]$ to be the subgroup of $G$ generated by the commutators of the form $$[h, k] = hkh^{-1}k^{-1} \qquad h\in H, k\in K.$$We know that $[H,K]=[K,H]$.

Lemma:

• Let $H$ and $K$ subgroups of $G$. Then $[H,K]\subseteq H$ iff $K\le N_G(H)$.
• If $K⊴G$ and $K\le H\le G$, then $[H,G]\le K$ iff $H/K\le Z(G/K)$.

Def: We define inductively a sequence of subgroups of $G$. $$L_1(G) := G, \qquad L_{n+1}(G) := [L_n(G), G]$$for $n\ge 2$. Just note that $L_2(G)=G^{(1)}$.

Prop: Let $G$ be a group. For any integer $n\ge 1$ the following are true:

• $L_{n+1}(G)\le L_n(G)$
• $L_n(G)$ is a normal subgroup of $G$.

This last proposition tells that the sequence of groups of $G$: $$G = : L_1(G) \supseteq L_2(G) \supseteq \dots \supseteq L_k(G) \supseteq \dots \supseteq {1} = 1$$is called the central inferior sequence of $G$.

Th: Let $G$ be a group. $Z^m(G)=G$ iff $L_{m+1}(G)$. It follows that $G$ is nilpotent of class $m$ iff $L_{m+1}(G)=1$, and $m$ is the least integer that satisfies this.