I have this question and I would like to know if my answer in particular "2-d)" is correct?.
Question:
Let $G$ be a finite group and let $p$ a prime divisor of $|G|=card(G)$ . In order to prove that $G$ has an element of order $p$ answer to the following question.
1- Let $E$ the set of all the familly $(x_1;...;x_p)$ of elements of $G$ such that $x_1 * x_2 * ... * x_p=e$. Show that the $ \sigma ((x_1;x_2;...;x_p)=(x_2;...;x_p;x_1) $ is a bijection of $E$ on itself. Compute $\sigma^p $.
2- Show that the natural action of $\langle\sigma \rangle$ on $E$ has strictly more than one fixed point and conclude.
Answer:
1-
$\sigma : E \rightarrow E $: Indeed $\forall \vec{x}=(x_1;...;x_p), \sigma ((x_1;...;x_p))=(x_2;...;x_p;x_1)$ and $(x_2;...;x_p;x_1)$ verifies $x_2 * ... * x_p * x_1=x_1^{-1} * x_1 * x_2 * ... * x_p * x_1=x_1^{-1} * e * x_1=e \in E$.
injectivity: $\forall \vec{x} \neq \vec{y} \Rightarrow \sigma(\vec{x})=(x_2;...;x_p;x_1) \neq (y_2;...;y_p;y_1)=\sigma (\vec{y})$
surjectivity: It is obvious that $\forall \vec{y}= (y_1;y_2;...;y_p;) ,\exists \vec{x'}=(x_1;x_2;...;x_p)=(y_p;y_1;y_2;...;y_{p-1}) \in E \Rightarrow \sigma(\vec{x'})=\vec{y}$
Is is trivial to show that $\sigma ^p(\vec{x})=\vec{x}$
2-
a)
It is trivial to show that $ (<\sigma>; \circ ) $ is a subgroup of $(S_p; \circ)$.
b)
There is at least one fixed point with $\sigma ^p = Id $, that gives for any element $x \in E \Rightarrow \sigma(x)=x$.
More over we know that by the definition of $E$ the element $\vec{x''}=(x_1=e;x_2=e;...;x_p=e) \in E$ and so in such a case no matter what $ \sigma^i $ we will take in $< \sigma > \Rightarrow \sigma^i(\vec{x''})=\vec{x''}$.
More generally all the element in $E$ of the form $(x_1=c; x_2=c; ...;x_p=c)$ (so that verify too $x_1 * x_2 * ... * x_p=e $ as they are in $E$) are too fixed points.
c)
Now let prove that $|E|$ is divisible by $p$.
By definition $E$ is build by choosing $p-1$ elements of $G$ as we want. That means that there are $|G|^{p-1}$ way of chosing the first $p-1$ elements. The last element (the $p-$th) has to verify (by def) $x_1 * ... * x_{p-1} * x_p = e $ and by definition of a group they can be only one such element like this in $G$ ( $x_p = (x_1 * ... * x_{p-1})^{-1}$). Hence $|E|=|G|^{p-1} = |G| \cdot ... \cdot |G|$ "$p-1$" times, so obviously $p$ divise $|E|$ too.
d)
Now if $(e;...; e) \in E$ where the only fixed points that will mean that there is exactly $|E|-1$ no fixed points.
Moreover by definition the orbit of a fixed point has only one element (trivial) and the orbit of no fixed point has stricly more than one element. More over because $p$ is prime each ot the no trivial orbit has exactly $p$ elements. Indeed if it was not the case (WLOG less than $p$) elements it will mean that $\exists k <p $ s.t. $ \sigma^k(x)=x$.
Each one of this $|E|-1$ orbits has $p$ element so $|E|-1$ is too divisible by $p$.
And this is a contradiction because $|E|$ too is divisible by $p$.
Q.E.D.
Again my question is (mostly) on the part "2-d)" is my reasonning correct?
Of course any feedback on the other parts of my answer will be welcome.
Thank you for your help.
According to https://math.stackexchange.com/users/1113334/amateur-algebraist my answer is ok.
I just wrotte this answer to mark my question as solved.