Exercise 3.2.9 (Cauchy's Theorem)

Proof. Let $G$ be a finite group of order $n$ and let $p$ be a prime dividing $n$, and let

(a)

Let’s calculate the cardinality of $\mathcal{}S$. Note that a $p$-tuple of $\mathcal{}S$ is determined by its first $p-1$ elements.

More precisely, let’s define $\phi ,\psi$ by:

$$\begin{array}{llll}\hfill & \phi \{\begin{array}{ccc}\hfill \mathcal{}S\hfill & \hfill \to \hfill & G^{p-1}\hfill \\ \hfill (x_1,\dots ,x_p)\hfill & \hfill \mapsto \hfill & (x_1,\cdots ,x_{p-1}),\hfill \end{array}& \hfill & \\ \hfill \text{and}& & \hfill \\ \hfill & \psi \{\begin{array}{ccc}\hfill G^{p-1}\hfill & \hfill \to \hfill & \mathcal{}S\hfill \\ \hfill (x_1,\dots ,x_{p-1})\hfill & \hfill \mapsto \hfill & (x_1,\dots ,x_{p-1},x_{p-1}^{-1}{x}_{p-2}^{-1}\cdots x_1^{-1}),\hfill \end{array}& \hfill & \end{array}$$ where $(x_1,\dots ,x_{p-1},x_{p-1}^{-1}{x}_{p-2}^{-1}\cdots x_1^{-1})\in \mathcal{}S$ because $x_1\cdots x_{p-1}(x_{p-1}^{-1}\cdots x_1^{-1})=1$.

Then $\phi$, $\psi$ satisfy $\phi \circ \psi ={\mathrm{id}}_{G^{p-1}}$ and $\psi \circ \phi ={\mathrm{id}}_{\mathcal{}S}$:

$$\begin{array}{llll}\hfill (\phi \circ \psi )(x_1,\dots ,x_{p-1})& =\phi (x_1,\dots ,x_{p-1},x_{p-1}^{-1}{x}_{p-2}^{-1}\cdots x_1^{-1})& \hfill & \\ \hfill & =(x_1,\dots ,x_{p-1})& \hfill & \end{array}$$

and since

$$\begin{array}{llll}\hfill (x_1,\dots ,x_p)\in \mathcal{}S& \Rightarrow x_1{x}_2\cdots x_p=1& \hfill & \\ \hfill & \Rightarrow x_p=x_{p-1}^{-1}{x}_{p-2}^{-1}\cdots x_1^{-1},& \hfill & \end{array}$$

we obtain

$$\begin{array}{llll}\hfill (\psi \circ \phi )(x_1,\dots ,x_p)& =\psi (x_1,\dots ,x_{p-1})& \hfill & \\ \hfill & =(x_1,\dots ,x_{p-1},x_{p-1}^{-1}{x}_{p-2}^{-1}\cdots x_1^{-1})& \hfill & \\ \hfill & =(x_1,x_2,\dots ,x_p).& \hfill & \end{array}$$

This shows that $\phi \circ \psi ={\mathrm{id}}_{G^{p-1}}$ and $\psi \circ \phi ={\mathrm{id}}_{\mathcal{}S}$, so $\phi$ is bijective.

Thus

$$|\mathcal{}S|=|G{\text{|}}^{p-1}.$$

(Therefore $p$ divides $|\mathcal{}S|$.)

(b)

If $x=(x_1,x_2,\dots ,x_p)\in G^p$ and $\sigma \in S_p$, we define $$\begin{array}{lll}\hfill x^{\sigma }& =(x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (p)}).& \hfill \text{(1)}\end{array}$$

This defines a right action of $S_p$ on $G^p$: if $\sigma ,{\sigma }^{\prime }\in S_p$, then

$$\begin{array}{llllll}\hfill {(x^{\sigma })}^{{\sigma }^{\prime }}& ={(x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (p)})}^{{\sigma }^{\prime }}& \hfill & & \hfill & \\ \hfill & ={(y_1,y_2,\dots ,y_p)}^{{\sigma }^{\prime }}& \hfill (\text{where }{y}_i=x_{\sigma (i)})& & \hfill & & \hfill \\ \hfill & =(y_{{\sigma }^{\prime }(1)},y_{{\sigma }^{\prime }(2)}\cdots ,y_{{\sigma }^{\prime }(p)})& \hfill & & \hfill & \\ \hfill & =(x_{(\sigma {\sigma }^{\prime })(1)},x_{(\sigma {\sigma }^{\prime })(2)},\dots ,x_{(\sigma {\sigma }^{\prime })(p)}& \hfill (\text{since }{y}_{{\sigma }^{\prime }(i)}=x_{\sigma ({\sigma }^{\prime }(i))})& & \hfill & & \hfill \\ \hfill & =x^{\sigma {\sigma }^{\prime }}.& \hfill & & \hfill & \\ \hfill & & \hfill & & \hfill \end{array}$$

In particular, if $\tau$ is the cyclic permutation $\tau =(12\cdots p)$, then

$$\begin{array}{llll}\hfill x^{\tau }& =(x_2,x_3,\cdots ,x_p,x_1).& \hfill & \end{array}$$

We first show that $x^{\tau }\in \mathcal{}S$.

In every group $G$, if if $a,b\in G$ satisfy $\mathit{ab}=1$, then $\mathit{ba}=b(\mathit{ab})b^{-1}=1$, so

$$\begin{array}{lll}\hfill \mathit{ab}=1\Rightarrow \mathit{ba}=1.& & \hfill \text{(2)}\end{array}$$

If $x=(x_1,x_2,\dots ,x_p)\in \mathcal{}S$, then $x_1{x}_2\cdots x_p=1$. We apply (2) to $a=x_1$ and $b=x_2\cdots x_p$, so that $\mathit{ab}=1$. We obtain $\mathit{ba}=1$, i.e.

$$x_2\cdots x_p{x}_1=1.$$

Therefore $x^{\tau }\in \mathcal{}S$. In conclusion,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in G^p,x\in \mathcal{}S\Rightarrow x^{\tau }\in \mathcal{}S.$$

Then for every integer $k\in \mathbb{Z}$,

$$x\in S\Rightarrow x^{{\tau }^k}\in \mathcal{}S.$$

This means that

$$x_1{x}_2\cdots x_p=x_2\cdots x_p{x}_1=x_3\cdots x_p{x}_1{x}_2=\cdots =x_p{x}_1{x}_2\cdots x_{p-1}.$$

( I interpret the ambiguous question "A cyclic permutation of an element of $\mathcal{}S$ is again an element of $\mathcal{}S$” as this result.)

(c)

By part (b), the group $H=⟨\tau ⟩=\{1,\tau ,\dots ,{\tau }^{p-1}\}$ acts to the right on $\mathcal{}S$, where $\tau =(12\dots p)$.

We define $\sim$ on $\mathcal{}S$ by

$$x\sim y⟺\exists <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},y=x^{{\tau }^k}(x,y\in \mathcal{}S),$$

that is

$$(x_1,x_2,\dots ,x_p)\sim (y_1,y_2,\dots ,y_p)⟺\exists <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},(y_1,y_2,\dots ,y_p)=(x_{{\tau }^k(1)},x_{{\tau }^k(2)},\dots ,x_{{\tau }^k(p)}).$$

This relation is the usual relation which expresses the fact that $x$ and $y$ are in the same orbit for the (right) action of $H$ on $\mathcal{}S$. We recall (see Prop. 2 p.114) that this is an equivalence relation:

• $x=x^{{\tau }^0}$, so $x\sim x$.

• If $x\sim y$, then there exists $k\in \mathbb{Z}$ such that $y=x^{{\tau }^k}$. Since $(x,\sigma )\mapsto x^{\sigma }$ is a right action of $H$ on $\mathcal{}S$,

  $$y^{{\tau }^{-k}}={(x^{{\tau }^k})}^{{\tau }^{-k}}=x^{{\tau }^k{\tau }^{-k}}=x^{{\tau }^0}=x,$$

  so $y\sim x$.

• If $x\sim y$ and $y\sim z$, then there are integers $k,l\in \mathbb{Z}$ such that $y=x^{{\tau }^k}$ and $z=y^{{\tau }^l}$. Using the same right action,

  $$z={(x^{{\tau }^k})}^{{\tau }^l}=x^{{\tau }^k{\tau }^l}=x^{{\tau }^{k+l}}(k+l\in \mathbb{Z}),$$

  so $x\sim z$.

  The class of $x=(x_1,x_2,\dots ,x_p)\in \mathcal{}S$ is the orbit of $x$ under the action of $H=⟨\tau ⟩$:

  $$\begin{array}{llll}\hfill \bar{x}& ={\mathcal{}O}_x& \hfill & \\ \hfill & =\{x,x^{\tau },x^{{\tau }^2},\cdots ,x^{{\tau }^{p-1}}\}& \hfill & \\ \hfill & =\{(x_1,x_2,\dots ,x_p),(x_2,x_3,\dots ,x_p,x_1),\dots ,(x_p,x_1,x_2,\dots ,x_{p-1})\}.& \hfill & \end{array}$$

(d)

Let $x\in \mathcal{}S$. If $x^{\tau }=x$, then $x^{{\tau }^2}={(x^{\tau })}^{\tau }=x^{\tau }=x$, and more generally $x^{{\tau }^k}=x$ for $k=0,1,\dots p-1$. So $$\begin{array}{llll}\hfill |{\mathcal{}O}_x|=1& ⟺x=x^{\tau }=x^{{\tau }^2}=\cdots =x^{{\tau }^{p-1}}& \hfill & \\ \hfill & ⟺x^{\tau }=x& \hfill & \\ \hfill & ⟺(x_2,x_3,\dots ,x_p,x_1)=(x_1,x_2,\dots ,x_p)& \hfill & \\ \hfill & ⟺x_1=x_2=\cdots =x_p& \hfill & \end{array}$$

So $|{\mathcal{}O}_x|=1$ if and only if $x=(a,a,\dots ,a)$ for some $a\in G$, and since $x\in \mathcal{}S$, $a^p=1$.

(e)

If $H_x$ is the stabilizer of $x\in S$ under the action of $H$, then (Proposition 2, section 4.1, p. 114) $$|{\mathcal{}O}_x|=(H:H_x)=|H|∕|H_x|,$$

so $|H_x|\cdot |{\mathcal{}O}_x|=|H|=p$. This shows that $|{\mathcal{}O}_x|$ is a divisor of $|H|=p$, where $p$ is a prime number, therefore

$$|{\mathcal{}O}_x|=1\text{ or }|{\mathcal{}O}_x|=p.$$

Let

$$Y=\{(x_1,x_2,\dots ,x_p)\in \mathcal{}S\mid x_1=x_2=\cdots =x_p\}.$$

By part (d), $Y$ is the set of elements of $x\in \mathcal{}S$ whose class contains a single element, so that $k=|Y|$ is the number of classes of size $1$, and let $d$ be the number of classes of size $p$.

Let $X$ be a complete system of representatives of the classes. Since the classes form a partition of $\mathcal{}S$,

$$\mathcal{}S=\underset{x\in X}{\bigcup }{\mathcal{}O}_x(\text{disjoint union}).$$

If an orbit is a class which contains a single element, i.e., ${\mathcal{}O}_x=\{x\}$, then its unique representative is $x$, thus $Y\subseteq X$, and

$$\begin{array}{llll}\hfill |\mathcal{}S|& =\underset{x\in X}{\sum }|{\mathcal{}O}_x|& \hfill & \\ \hfill & =\underset{x\in Y}{\sum }|{\mathcal{}O}_x|+\underset{x\in X\setminus Y}{\sum }|{\mathcal{}O}_x|& \hfill & \\ \hfill & =\underset{x\in Y}{\sum }1+\underset{x\in X\setminus Y}{\sum }|{\mathcal{}O}_x|& \hfill & \\ \hfill & =|Y|+\underset{x\in X\setminus Y}{\sum }|{\mathcal{}O}_x|& \hfill & \\ \hfill & =k+\mathit{pd},& \hfill & \end{array}$$

because every orbit ${\mathcal{}O}_x$, where $x\in Y\setminus X$, has size $p$, and there are $d$ such classes.

(f)

Since the class of $(1,1,\dots ,1)$ has size $1$, $Y\ne \varnothing$, so $|Y|\ge 1$.

By part (e), $|\mathcal{}S|=p^{p-1}=k+\mathit{pd}$, thus $p\mid k=|Y|$, where $|Y|\ge 1$. Therefore $|Y|>1$, so $Y$ contains an element $(a,a,\dots a)$, where $a\in G$, $a\ne 1$. Then $(a,a,\dots ,a)\in \mathcal{}S$, therefore

$$a^p=1,a\ne 1.$$

This shows that the order of $a$ is $p$:

Every group $G$ such that $p$ divides $|G|$ contains an element of order $p$ (Cauchy’s Theorem).