properties of group commutators and commutator subgroups

The purpose of this entry is to collect properties of http://planetmath.org/node/2812groupcommutators and commutator subgroups. Feel free to add more theorems!

Let $G$ be a group.

Theorem 1.

Let $x\mathrm{,}y\mathrm{\in }G$, then ${\mathrm{[}x\mathrm{,}y\mathrm{]}}^{\mathrm{-}\mathrm{1}}\mathrm{=}\mathrm{[}y\mathrm{,}x\mathrm{]}$.

Proof.

Direct computation yields

$${[x,y]}^{-1}={(x^{-1}{y}^{-1}xy)}^{-1}=y^{-1}{x}^{-1}yx=[y,x].$$

∎

Theorem 2.

Let $X\mathrm{,}Y$ be subsets of $G$, then $\mathrm{[}X\mathrm{,}Y\mathrm{]}\mathrm{=}\mathrm{[}Y\mathrm{,}X\mathrm{]}$.

Proof.

By Theorem 1, the elements from $[X,Y]$ or$[Y,X]$ are products of commutators of the form $[x,y]$ or $[y,x]$with $x\in X$ and $y\in Y$.∎

Theorem 3 (Hall–Witt identity).

Let $x\mathrm{,}y\mathrm{,}z\mathrm{\in }G$, then

$$y^{-1}[x,y^{-1},z]y{z}^{-1}[y,z^{-1},x]z{x}^{-1}[z,x^{-1},y]x=1.$$

Proof.

This is mainly a brute-force calculation. We can easily calculate thefirst factor $y^{-1}[x,y^{-1},z]y$ explicitly usingtheorem 1:

$y^{-1}[x,y^{-1},z]y$

$=$$y^{-1}[y^{-1},x]z^{-1}[x,y^{-1}]zy$

$=$$y^{-1}y{x}^{-1}{y}^{-1}x{z}^{-1}{x}^{-1}yx{y}^{-1}zy$

$=$$x^{-1}{y}^{-1}x{z}^{-1}{x}^{-1}yx{y}^{-1}zy.$

Let $h_1:=x^{-1}{y}^{-1}x{z}^{-1}{x}^{-1}$, the “first half” of$y^{-1}[x,y^{-1},z]y$. Let $h_2$ be the element obtained from $h_1$ bythe cyclic shift $S:x\mapsto y\mapsto z\mapsto x$, and $h_3$ bethe element obtained from $h_2$ by $S$. We have

$$h_2^{-1}={(y^{-1}{z}^{-1}y{x}^{-1}{y}^{-1})}^{-1}=yx{y}^{-1}zy$$

which gives us

$$y^{-1}[x,y^{-1},z]y=h_1{h}_2^{-1},$$

and, by applying $S$ twice

$z^{-1}[y,z^{-1},x]z$$=h_2{h}_3^{-1},$

$x^{-1}[z,x^{-1},y]x$$=h_3{h}_1^{-1}.$

In total, we have

$$y^{-1}[x,y^{-1},z]y{z}^{-1}[y,z^{-1},x]z{x}^{-1}[z,x^{-1},y]x=h_1{h}_2^{-1}{h}_2{h}_3^{-1}{h}_3{h}_1^{-1}=1.$$

∎

Theorem 4 (Three subgroup lemma).

Let $N$ be a normal subgroup of $G$. Furthermore, let $X$, $Y$ and $Z$be subgroups of $G$, such that $\mathrm{[}X\mathrm{,}Y\mathrm{,}Z\mathrm{]}$ and $\mathrm{[}Y\mathrm{,}Z\mathrm{,}X\mathrm{]}$ are containedin $N$. Then $\mathrm{[}Z\mathrm{,}X\mathrm{,}Y\mathrm{]}$ is contained in $N$ as well.

Proof.

The group $[Z,X,Y]$ is generated by all elements of the form$[z,x^{-1},y]$ with $x\in X$, $y\in Y$ and $z\in Z$. Since $N$ isnormal, $y^{-1}[x,y^{-1},z]y$ and $x^{-1}[z,x^{-1},y]x$ are elementsof $N$. The Hall–Witt identity then implies that$x^{-1}[z,x^{-1},y]x$ is an element of $N$ as well. Again, since $N$is normal, $[z,x^{-1},y]\in N$ which concludes the proof.∎

Theorem 5.

For any $x\mathrm{,}y\mathrm{,}z\mathrm{\in }G$ we have

$[xy,z]$$=$${[x,z]}^y[y,z]$

$[x,yz]$$=$$[x,z]{[x,y]}^z$

${[x,y]}^z$$=$$[x^z,y^z]$

$[x^z,y]$$=$$[x,y^{z^{-1}}]$

where $a^b$ denotes $b^{\mathrm{-}\mathrm{1}}\mathit{}a\mathit{}b$

Proof.

By expanding:

$[xy,z]$$=$$y^{-1}{x}^{-1}{z}^{-1}xyz$

$=$$y^{-1}{x}^{-1}{z}^{-1}\cdot xz\cdot z^{-1}{x}^{-1}\cdot xyz$

$=$$y^{-1}[x,z]\cdot y\cdot y^{-1}\cdot z^{-1}{x}^{-1}\cdot xyz$

$=$${[x,z]}^y\cdot y^{-1}{z}^{-1}yz$

$=$${[x,z]}^y[y,z]$

The other identities are proved similarly.∎