# Cauchy's theorem proof clarification (group theory)

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.

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 $$\;(x,x,...,x)\;$$ .
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 $$(e,e,\ldots,e)$$.
$$(g_0,\dots,g_{p-1})$$ 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.