Cauchy's theorem proof clarification (group theory)

Question

Cauchy's theorem says that if $G$ is a finite group with $p | |G|$ (when $p$ is prime), then $G$ contains an element of order $p$.

When following the proof from wikipedia: https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory) (proof 2),

They conclude that there exists an orbit $O(x)$ of size $1$ such that $x\neq (e,e,...,e)$. But why that means that $x = (a, a, ..., a)$ for some $a\in G$? Why can't $x$ be $x = (g_0, ..., g_{p-1})$ for different $g_i$?

Thanks.

Answer 1

Take an element $\;(x_1,...,x_p)\in X\;$ . If its orbit has only one element, it means that if $\;C_p=\langle z\rangle\;$ then $$(x_1,...,x_p)=z(x_1,...,x_p)=(x_2,...,x_p,x_1)=z^2(x_1,...,x_p)=(x_3,...,x_p,x_1,x_2)=\ldots$$

and we get that $\;x_1=x_2=\ldots=x_p\;$, so the element is in fact of the form $\;(x,x,...,x)\;$ .

Answer 2

$(g_0,\dots,g_{p-1})$ would be an orbit of size $p$. What's meant by an orbit of size $1$ is that there's only $1$ element.

The action is cyclic permutation of the components. An orbit of size one then means an element of the form $(a,a,\dots,a)$. Permuting the components can then, and only then, result in nothing different.

Answer 3

What they prove here is that there are two types of orbits: the orbits of the elements of the type $(g,g,\ldots,g)$ (which have a single element) and all the other orbits. Then they prove that the number of orbits of the first type must be a multiple of $p$. So, there is some other orbit besides $(e,e,\ldots,e)$.