Suppose $G$ is a group and $|G|=p^k$ , and $X$ is finite set. Let $G$ act on $X$ or the cardinality $|X|$

Question

While studying group theory I came across this question:

Suppose $G$ is a group and $|G|=p^k$ for some $p$ prime number and $k$ integer, and $X$ is finite set. Let $G$ act on $X$ if $p$ does not divide $|X|$ see if $G_{\alpha}=G$ for some $\alpha \in X$ true or not.

I need a hint the relation between the cardinality if $X$ and order of the group $G$. In other words why does the number of elements in $X$ matter?

Answer 1

Hint:

Use the orbit-stabilizer theorem.

If $G$ acts on $X$, then every point $x \in X$ has some orbit. Moreover, since every element is in exactly one orbit we see

$$|X| = \sum_{\text{orbits}} |\text{orbit}|$$

By the orbit-stabilizer theorem, though, we know each orbit satisfies $|\text{orbit}| \cdot |\text{stabilizer}| = |G|$.

If $|X|$ is not a multiple of $p$, but each $|\text{orbit}|$ must divide $|G| = p^k$, what does this tell you about the size of some orbits? It might be helpful to work mod $p$.

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I hope this helps ^_^