IMC 2020 · Day 2 · P7

Denote $K = H_1 \cap H_2$. Since $$n (n - 1) = [G : K] = [G : H_1][H_1 : K] = n [H_1 : K],$$ we obtain that $[H_1 : K] = n - 1$. Thus, the subgroup $H_1$ is partitioned into $n - 1$ left cosets of $K$, say $H_1 = \bigsqcup_{i=1}^{n-1} h_i K$. Therefore, the set $H_1 H_2 = \{ ab : a \in H_1, b \in H_2 \}$ is partitioned as $$H_1 H_2 = \left( \bigsqcup_{i=1}^{n-1} h_i K \right) H_2 = \bigsqcup_{i=1}^{n-1} h_i K H_2 = \bigsqcup_{i=1}^{n-1} h_i H_2.$$ The last equality holds because $K \subseteq H_2$, so $K H_2 = H_2$. The last expression is a disjoint union since $$h_i H_2 \cap h_j H_2 \ne \emptyset \iff h_i^{-1} h_j \in H_2 \iff h_i^{-1} h_j \in K \iff h_i = h_j.$$ Thus, $H_1 H_2$ is a disjoint union of $n - 1$ left cosets with respect to $H_2$; hence $L = G \setminus (H_1 H_2)$ is the remaining such left coset. Similarly, $L$ is a right coset with respect to $H_1$. Therefore, for each $g \in L$ we have $L = g H_2 = H_1 g$, which yields $H_2 = g^{-1} H_1 g$.

Put $G / H_1 = X$ and $G / H_2 = Y$, those are $n$-element sets acted on by $G$ from the left. Let $G$ act on $X \times Y$ from the left coordinate-wise, consider this product as a table, with rows being copies of $X$ and columns being copies of $Y$.

The stabilizer of a point $(x, y)$ in $X \times Y$ is $H_1 \cap H_2$. By the orbit-stabilizer theorem, we obtain that the orbit $Z$ of $(x, y)$ has size $[G : H_1 \cap H_2] = n (n - 1)$.

If $Z$ contains a whole column then there is a subgroup $G_1$ of $G$ that stabilizes $x$ and acts transitively on $Y$. If we conjugate $G_1$ to a group $G'_1$, then its action remains transitive on $Y$, so by conjugation we obtain columns of the table. Since $G$ acts transitively on $X$, we cover all the columns. It follows that $Z = X \times Y$, so $$n (n - 1) = |Z| = |X \times Y| = n^2,$$ which is a contradiction.

Hence every column of $X \times Y$ has an element not from $Z$. The same holds for the rows of $X \times Y$. There are $n$ elements not from $Z$ in total and they induce a bijection between $X$ and $Y$ which allows us to identify $X = Y$.

After this identification, every element $(x, x)$ from the diagonal of $X \times X$ (i.e. from $(X \times X) \setminus Z$) is moved to a diagonal element by any $g \in G$, because $g x = g x$. In this formula the action of $g$ in the left hand side and the action of $g$ in the right hand side are the actions of $g$ on $X$ and $Y$ respectively.

Therefore our bijection between $X$ and $Y$ is an isomorphism of sets with a left action of $G$. Since $H_1$ and $H_2$ are stabilizers of the points in the same transitive action of $G$, we conclude that they are conjugate.

Remark. The situation in the problem is possible for every $n \ge 2$: let $G = S_n$ and let $H_1$ an $H_2$ be the stabilizer subgroups of two elements.