If $G$ is a group that acts on a finite set $X$, prove that there exists an integer$ N>0$, such that $g^{N}x=x$ for all $x\in X$

Question

If $G$ is a group that acts on a finite set $X$, prove that there exists an integer$ N>0$, such that $g^{N}x=x$ for all $x\in X$

I'm having a lot of trouble trying to figure out how to write this proof for my course. I don't really understand the concept of groups acting on sets, so I'm trying to review my notes and textbook but it's not really clicking. I was given the hint that the action of $G$ on $X$ can be seen as a morphism from $G$ to $Perm(X)$, the permutation group of $X$. So the action of a fixed $g \in G$ permutes the elements in $X$.

In my notes I have that if a group $G$ acts on a set $X$, then it satisfies:

• $e_gx=x \forall x \in X$
• $(gh)x=g(hx), \forall g,h \in G, \forall x\in X$

I also have that one property of this is:

• $T:G\rightarrow Perm(X)=${$f:X\rightarrow X$ | $f$ bijective}, $T(g)(x)=T_g(x)=gx$, T is a morphism of groups.

I know that a homomorphism of groups preserves identities such that $T(e_G)=e_X$, so I'm assuming that I want to either show that

• $g^N=e_G$ such that $g^Nx=x \iff e_Gx=x$.

Or

• The action of $g^N \in G$ on $x \in X$ permutes the element $x$ in such a way that we get $e_Xx$ or simply $x \in X$.

I'm trying to connect the dots, but am not sure if I'm even working with the correct dots. Any sort of guidance or helpful insight would be incredibly appreciated!

Answer 1

Since $X$ is a finite set, suppose $X$ contains $n$ distinct elements $X=\{x_1, x_2,...,x_n\}$ and consider the set $Y$ of all permutations on n-tuples $(x_1, x_2,...,x_n)$. Define: $$g(x_1, x_2,...,x_n)=(gx_1, gx_2,...,gx_n)$$

It is easy to check:

$(g_1g_2)(x_1, x_2,...,x_n)=g_1(g_2(x_1, x_2,...,x_n))$ and $e(x_1, x_2,...,x_n)=(x_1, x_2,...,x_n)$

So, $Y$ is a $G$-set. Now, for $\forall g\in G$, since $|Y|=n!$, if we consider the following elements $(e=g^0)$:

$g^0(x_1, x_2,...,x_n), ~~g^1(x_1, x_2,...,x_n), ~~g^2(x_1, x_2,...,x_n)..., ~~g^{n!}(x_1, x_2,...,x_n)$

There are totally $n!+1$ elements but at most $n!$ of them are distinct, which means there must be at least two of them are identical. Therefore, $\exists 0\le k_1< k_2\le n!$, such that $g^{k_1}(x_1, x_2,...,x_n)=g^{k_2}(x_1, x_2,...,x_n)$, which gives:

$$g^{k_2-k_1}(x_1, x_2,...,x_n)=(x_1, x_2,...,x_n)$$

Further,

$$g^{k_2-k_1}x_i=x_i, ~~i=1,2,...,n$$

Now we can let $N=k_2-k_1$, where $1\le N\le n!$, such that

$$g^{N}x=x, \forall x\in X$$