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Proving Cauchy's Theorem in Group Theory

  • #1
1,456
44

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.
 
Last edited:

Answers and Replies

  • #2
12,691
9,235

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##.
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.
 
  • #3
1,456
44
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##.
 

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