- #1

Mr Davis 97

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## 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).

$$

## Homework Equations

## 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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