# Proving Cauchy's Theorem in Group Theory

## Homework Statement

Let $S = \{(x_1, \dots, x_p) \mid x_i \in G, x_1 x_2 \cdots x_p = e\}$.

Let $C_p$ denote cyclic subgroup of $S_p$ of order $p$ generated by the $p$-cycle, $\sigma = (1 \, 2 \, \cdots \, p)$. Show that the following rule gives an action of $C_p$ on $S$
$$\sigma \cdot (x_1, \dots, x_p) := (x_{\sigma(1)}, \dots, x_{\sigma(p)}) = (x_2, x_3, \dots, x_p, x_1).$$

## The Attempt at a Solution

In general showing that something is an action is not too difficult, since we only have to check two things. However, I am a but confused with how this action is defined. Normally, when group actions are defined they are defined for arbitrary elements of the group $G$. But here it is only defined for the generator of $G$. So when I try to prove that this is a group action, do I say: Suppose $\alpha, \beta \in C_p$. Then $\alpha=\sigma^a$ and $\beta=\sigma^b$ for some integers $a,b\in [0, p)$. Basically, does the following proof work?

1) Let $\sigma^i,\sigma^j\in C_p$. Then
\begin{align*} \sigma^i\cdot (\sigma^j \cdot (x_1,\dots,x_p)) &= \sigma^i\cdot (x_{1+j},\cdots,x_{p+j})\\ &=(x_{(1+j)+i},\dots,x_{(p+j)+i})\\ &=(x_{1+(j+i)},\dots,x_{p+(j+i)})\\ &=\sigma^{i+j}\cdot (x_1,\dots,x_p)\\ &=(\sigma^{i}\sigma^j)\cdot (x_1,\dots,x_p) \end{align*}

2) Clearly $\operatorname{id}_{C_p}\cdot (x_1,\dots,x_p) = (x_1,\dots,x_p)$.

So we have a group action.

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Mentor

## Homework Statement

Let $S = \{(x_1, \dots, x_p) \mid x_i \in G, x_1 x_2 \cdots x_p = e\}$.

Let $C_p$ denote cyclic subgroup of $S_p$ of order $p$ generated by the $p$-cycle, $\sigma = (1 \, 2 \, \cdots \, p)$. Show that the following rule gives an action of $C_p$ on $S$
$$\sigma \cdot (x_1, \dots, x_p) := (x_{\sigma(1)}, \dots, x_{\sigma(p)}) = (x_2, x_3, \dots, x_p, x_1).$$

## The Attempt at a Solution

In general showing that something is an action is not too difficult, since we only have to check two things. However, I am a but confused with how this action is defined. Normally, when group actions are defined they are defined for arbitrary elements of the group $G$. But here it is only defined for the generator of $G$.
That's not how I read it. First of all, we do not have generators, only simple group elements. So I read $S$ as a subset of $G^p$ such that its "digit product" is the neutral element. For a group action, we only need a set.
So when I try to prove that this is a group action, do I say: Suppose $\alpha, \beta \in C_p$. Then $\alpha=\sigma^a$ and $\beta=\sigma^b$ for some integers $a,b\in [0, p)$. Basically, does the following proof work?

1) Let $\sigma^i,\sigma^j\in C_p$. Then
\begin{align*} \sigma^i\cdot (\sigma^j \cdot (x_1,\dots,x_p)) &= \sigma^i\cdot (x_{1+j},\cdots,x_{p+j})\\ &=(x_{(1+j)+i},\dots,x_{(p+j)+i})\\ &=(x_{1+(j+i)},\dots,x_{p+(j+i)})\\ &=\sigma^{i+j}\cdot (x_1,\dots,x_p)\\ &=(\sigma^{i}\sigma^j)\cdot (x_1,\dots,x_p) \end{align*}

2) Clearly $\operatorname{id}_{C_p}\cdot (x_1,\dots,x_p) = (x_1,\dots,x_p)$.

So we have a group action.
My concern is less the action itself, but how do we know that all permutations multiply to $e$ again, i.e. is $S$ closed under this operation? This part is missing in your argument.

That's not how I read it. First of all, we do not have generators, only simple group elements. So I read $S$ as a subset of $G^p$ such that its "digit product" is the neutral element. For a group action, we only need a set.

My concern is less the action itself, but how do we know that all permutations multiply to $e$ again, i.e. is $S$ closed under this operation? This part is missing in your argument.
Well, note that in general for groups, if $a,b\in G$ and if $ab=e$ then $ba=e$. So if we have $(x_1,\dots,x_p)$ with $x_1\cdots x_p=e$, then with $(x_2,x_3\dots,x_p,x_1)$ we have $(x_2x_3\cdots x_p)(x_1) = (x_1)(x_2x_3\cdots x_p) = e$.

• fresh_42