IMC 2010 · Day 2 · P8

Let us consider the action of $G$ on the set $X = \{1, \dots, n\}$. Let $$G_x = \{ g \in G : g(x) = x \} \quad \text{and} \quad Gx = \{ g(x) : g \in G \}$$ be the stabilizer and the orbit of $x \in X$ under this action, respectively. The condition of the problem states that $$\tag{1} G = \bigcup_{x \in X} G_x$$ and $$\tag{2} G_x \cap G_y = \{e\} \quad \text{for all } x \ne y.$$ We need to prove that $G = G_x$ for some $x \in X$.

Let $G x_1, \dots, G x_k$ be the distinct orbits of the action of $G$. Then one can write (1) as $$\tag{3} G = \bigcup_{i=1}^{k} \bigcup_{y \in G x_i} G_y.$$ It is well known that $$\tag{4} |Gx| = \frac{|G|}{|G_x|}.$$ Also note that if $y \in Gx$ then $Gy = Gx$ and thus $|Gy| = |Gx|$. Therefore, $$\tag{5} |G_x| = \frac{|G|}{|Gx|} = \frac{|G|}{|Gy|} = |G_y| \quad \text{for all } y \in Gx.$$ Combining (3), (2), (4) and (5) we get $$|G| - 1 = |G \setminus \{e\}| = \left| \bigcup_{i=1}^{k} \bigcup_{y \in G x_i} G_y \setminus \{e\} \right| = \sum_{i=1}^{k} \frac{|G|}{|G_{x_i}|} (|G_{x_i}| - 1),$$ hence $$\tag{6} 1 - \frac{1}{|G|} = \sum_{i=1}^{k} \left( 1 - \frac{1}{|G_{x_i}|} \right).$$ If for some $i, j \in \{1, \dots, k\}$ $|G_{x_i}|, |G_{x_j}| \ge 2$ then $$\sum_{i=1}^{k} \left( 1 - \frac{1}{|G_{x_i}|} \right) \ge \left( 1 - \frac{1}{2} \right) + \left( 1 - \frac{1}{2} \right) = 1 > 1 - \frac{1}{|G|}$$ which contradicts with (6), thus we can assume that $$|G_{x_1}| = \dots = |G_{x_{k-1}}| = 1.$$ Then from (6) we get $|G_{x_k}| = |G|$, hence $G_{x_k} = G$.