| Version 4 |
Version 3 |
| \newtheorem{thm}{Theorem} |
\newtheorem{thm}{Theorem} |
| The purpose of this entry is to collect properties of \PMlinkid{group |
The purpose of this entry is to collect properties of \PMlinkid{group |
| commutators}{2812} and commutator subgroups. Feel free to add more theorems! |
commutators}{2812} and commutator subgroups. Feel free to add more theorems! |
|
|
| Let $G$ be a group. |
Let $G$ be a group. |
|
|
| \begin{thm} |
\begin{thm} |
| \label{thm:inversecommutator} |
\label{thm:inversecommutator} |
| Let $x,y\in G$, then $[x,y]^{-1}=[y,x]$. |
Let $x,y\in G$, then $[x,y]^{-1}=[y,x]$. |
| \end{thm} |
\end{thm} |
| \begin{proof} |
\begin{proof} |
| Direct computation yields |
Direct computation yields |
| \begin{equation*} |
\begin{equation*} |
| [x,y]^{-1}=(x^{-1}y^{-1}xy)^{-1}=y^{-1}x^{-1}yx=[y,x]. |
[x,y]^{-1}=(x^{-1}y^{-1}xy)^{-1}=y^{-1}x^{-1}yx=[y,x]. |
| \end{equation*} |
\end{equation*} |
| \end{proof} |
\end{proof} |
|
|
| \begin{thm} |
\begin{thm} |
| Let $X,Y$ be subsets of $G$, then $[X,Y]=[Y,X]$. |
Let $X,Y$ be subsets of $G$, then $[X,Y]=[Y,X]$. |
| \end{thm} |
\end{thm} |
| \begin{proof} |
\begin{proof} |
| By Theorem~\ref{thm:inversecommutator}, the elements from $[X,Y]$ or |
By Theorem~\ref{thm:inversecommutator}, the elements from $[X,Y]$ or |
| $[Y,X]$ are products of commutators of the form $[x,y]$ or $[y,x]$ |
$[Y,X]$ are products of commutators of the form $[x,y]$ or $[y,x]$ |
| with $x\in X$ and $y\in Y$. |
with $x\in X$ and $y\in Y$. |
| \end{proof} |
\end{proof} |
|
|
| \begin{thm}[Hall--Witt identity] |
\begin{thm}[Hall--Witt identity] |
| Let $x,y,z\in G$, then |
Let $x,y,z\in G$, then |
| \begin{equation*} |
\begin{equation*} |
| y^{-1}[x,y^{-1},z]yz^{-1}[y,z^{-1},x]zx^{-1}[z,x^{-1},y]x=1. |
y^{-1}[x,y^{-1},z]yz^{-1}[y,z^{-1},x]zx^{-1}[z,x^{-1},y]x=1. |
| \end{equation*} |
\end{equation*} |
| \end{thm} |
\end{thm} |
| \begin{proof} |
\begin{proof} |
| This is mainly a brute-force calculation. We can easily calculate the |
This is mainly a brute-force calculation. We can easily calculate the |
| first factor $y^{-1}[x,y^{-1},z]y$ explicitly using |
first factor $y^{-1}[x,y^{-1},z]y$ explicitly using |
| theorem~\ref{thm:inversecommutator}: |
theorem~\ref{thm:inversecommutator}: |
| \begin{align*} |
\begin{align*} |
| &y^{-1}[x,y^{-1},z]y\\ |
&y^{-1}[x,y^{-1},z]y\\ |
| =&y^{-1}[y^{-1},x]z^{-1}[x,y^{-1}]zy\\ |
=&y^{-1}[y^{-1},x]z^{-1}[x,y^{-1}]zy\\ |
| =&y^{-1}yx^{-1}y^{-1}xz^{-1}x^{-1}yxy^{-1}zy\\ |
=&y^{-1}yx^{-1}y^{-1}xz^{-1}x^{-1}yxy^{-1}zy\\ |
| =&x^{-1}y^{-1}xz^{-1}x^{-1}yxy^{-1}zy. |
=&x^{-1}y^{-1}xz^{-1}x^{-1}yxy^{-1}zy. |
| \end{align*} |
\end{align*} |
| Let $h_1:=x^{-1}y^{-1}xz^{-1}x^{-1}$, the ``first half'' of |
Let $h_1:=x^{-1}y^{-1}xz^{-1}x^{-1}$, the ``first half'' of |
| $y^{-1}[x,y^{-1},z]y$. Let $h_2$ be the element obtained from $h_1$ by |
$y^{-1}[x,y^{-1},z]y$. Let $h_2$ be the element obtained from $h_1$ by |
| the cyclic shift $S\colon x\mapsto y\mapsto z\mapsto x$, and $h_3$ be |
the cyclic shift $S\colon x\mapsto y\mapsto z\mapsto x$, and $h_3$ be |
| the element obtained from $h_2$ by $S$. We have |
the element obtained from $h_2$ by $S$. We have |
| \begin{equation*} |
\begin{equation*} |
| h_2^{-1}=(y^{-1}z^{-1}yx^{-1}y^{-1})^{-1}=yxy^{-1}zy |
h_2^{-1}=(y^{-1}z^{-1}yx^{-1}y^{-1})^{-1}=yxy^{-1}zy |
| \end{equation*} |
\end{equation*} |
| which gives us |
which gives us |
| \begin{equation*} |
\begin{equation*} |
| y^{-1}[x,y^{-1},z]y=h_1h_2^{-1}, |
y^{-1}[x,y^{-1},z]y=h_1h_2^{-1}, |
| \end{equation*} |
\end{equation*} |
| and, by applying $S$ twice |
and, by applying $S$ twice |
| \begin{align*} |
\begin{align*} |
| z^{-1}[y,z^{-1},x]z&=h_2h_3^{-1},\\ |
z^{-1}[y,z^{-1},x]z&=h_2h_3^{-1},\\ |
| x^{-1}[z,x^{-1},y]x&=h_3h_1^{-1}. |
x^{-1}[z,x^{-1},y]x&=h_3h_1^{-1}. |
| \end{align*} |
\end{align*} |
| In total, we have |
In total, we have |
| \begin{equation*} |
\begin{equation*} |
| y^{-1}[x,y^{-1},z]yz^{-1}[y,z^{-1},x]zx^{-1}[z,x^{-1},y]x=h_1h_2^{-1}h_2h_3^{-1}h_3h_1^{-1}=1. |
y^{-1}[x,y^{-1},z]yz^{-1}[y,z^{-1},x]zx^{-1}[z,x^{-1},y]x=h_1h_2^{-1}h_2h_3^{-1}h_3h_1^{-1}=1. |
| \end{equation*} |
\end{equation*} |
| \end{proof} |
\end{proof} |
|
|
| \begin{thm}[Three subgroup lemma] |
\begin{thm}[Three subgroup lemma] |
| Let $N$ be a normal subgroup of $G$. Furthermore, let $X$, $Y$ and $Z$ |
Let $N$ be a normal subgroup of $G$. Furthermore, let $X$, $Y$ and $Z$ |
| be subgroups of $G$, such that $[X,Y,Z]$ and $[Y,Z,X]$ are contained |
be subgroups of $G$, such that $[X,Y,Z]$ and $[Y,Z,X]$ are contained |
| in $N$. Then $[Z,X,Y]$ is contained in $N$ as well. |
in $N$. Then $[Z,X,Y]$ is contained in $N$ as well. |
| \end{thm} |
\end{thm} |
| \begin{proof} |
\begin{proof} |
| The group $[Z,X,Y]$ is generated by all elements of the form |
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 $N$ is |
$[z,x^{-1},y]$ with $x\in X$, $y\in Y$ and $z\in Z$. Since $N$ is |
| normal, $y^{-1}[x,y^{-1},z]y$ and $x^{-1}[z,x^{-1},y]x$ are elements |
normal, $y^{-1}[x,y^{-1},z]y$ and $x^{-1}[z,x^{-1},y]x$ are elements |
| of $N$. The Hall--Witt identity then implies that |
of $N$. The Hall--Witt identity then implies that |
| $x^{-1}[z,x^{-1},y]x$ is an element of $N$ as well. Again, since $N$ |
$x^{-1}[z,x^{-1},y]x$ is an element of $N$ as well. Again, since $N$ |
| is normal, $[z,x^{-1},y]\in N$ which concludes the proof. |
is normal, $[z,x^{-1},y]\in N$ which concludes the proof. |
| \end{proof} |
\end{proof} |
