Number of elements of a set via action

Question

Let $G$ be a finite group, acting on a finite set $X$. Burnside's lemma says that the following holds $$|X/G| = \frac{1}{|G|}\sum_{g\in G}|St(g)|$$ I'm wondering does it mean that $|X| = |X/G|\cdot|G|$ as in Lagrange's theorem for groups. Or maybe there's a classification of actions for which this is true.

Answer 1

Edit: I am assuming that $X$ and $G$ are finite for everything to make sense.

Note that precisely by Burnside's lemma, your question translates to asking wether

$$ |X| = \sum_{g \in G}|X^g| $$

where $X^g := \{x \in X : g \cdot x = x\}$.

However, when $g = 1$ we have

$$ X^1 = \{x \in X : 1 \cdot x = x\} = X $$

so in turn this would mean that

$$ 0 = \sum_{g \neq 1}|X^g|, $$

or equivalently, that $|X^g| = 0$ for all $g \neq 1$.

Thus equality holds when $g \cdot x \neq x $ for all $x \in X, g \neq 1$. If I recall correctly, this is the definition of a free action.