If a Finite Group Acts on a Set Freely and Transitively, then the Numbers of Elements are the Same

Problem 488

Let $G$ be a finite group and let $S$ be a non-empty set.
Suppose that $G$ acts on $S$ freely and transitively.
Prove that $|G|=|S|$. That is, the number of elements in $G$ and $S$ are the same.

A group action of a group $G$ on a set $S$ is called free if whenever we have
\[gs=hs\]
for some $g, h\in G$ and $s\in S$, this implies $g=h$.

A group action of a group $G$ on a set $S$ is called transitive if for each pair $s, t\in S$ there exists an element $g\in G$ such that
\[gs=t.\]

Proof.

We simply denote by $gs$ the action of $g\in G$ on $s\in S$.

Since $S$ is non-empty, we fix an element $s_0 \in S$. Define a map
\[\phi: G \to S\]
by sending $g\in G$ to $gs_0 \in S$.
We prove that the map $\phi$ is bijective.

Suppose that we have $\phi(g)=\phi(h)$ for some $g, h\in G$.
Then it gives $gs_0=hs_0$, and since the action is free this implies that $g=h$.
Thus $\phi$ is injective.

To show that $\phi$ is surjective, let $s$ be an arbitrary element in $S$.
Since the action is transitive, there exists $g\in G$ such that $gs_0=s$.
Hence we have $\phi(g)=s$, and $\phi$ is surjective.

Therefore the map $\phi:G\to S$ is bijective, and we conclude that $|G|=|S|$.

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