International Mathematics Competition
for University Students
2020

IMC2020: Day 2, Problem 7

Problem 7. Let \(\displaystyle G\) be a group and \(\displaystyle n\ge2\) be an integer. Let \(\displaystyle H_1\) and \(\displaystyle H_2\) be two subgroups of \(\displaystyle G\) that satisfy

\(\displaystyle [G:H_1]=[G:H_2]=n \quad\text{and}\quad [G:(H_1\cap H_2)]=n(n-1). \)

Prove that \(\displaystyle H_1\) and \(\displaystyle H_2\) are conjugate in \(\displaystyle G\).

(Here \(\displaystyle [G:H]\) denotes the index of the subgroup \(\displaystyle H\), i.e. the number of distinct left cosets \(\displaystyle xH\) of \(\displaystyle H\) in \(\displaystyle G\). The subgroups \(\displaystyle H_1\) and \(\displaystyle H_2\) are conjugate if there exists an element \(\displaystyle g\in G\) such that \(\displaystyle g^{-1}H_1g=H_2\).)

Ilya Bogdanov and Alexander Matushkin, Moscow Institute of Physics and Technology

Solution 1. Denote \(\displaystyle K=H_1\cap H_2\). Since

\(\displaystyle n(n-1)=[G:K]=[G:H_1][H_1:K]=n[H_1:K],\)

we obtain that \(\displaystyle [H_1:K]=n-1\). Thus, the subgroup \(\displaystyle H_1\) is partitioned into \(\displaystyle n-1\) left cosets of \(\displaystyle K\), say \(\displaystyle H_1=\bigsqcup_{i=1}^{n-1} h_iK\). Therefore, the set \(\displaystyle H_1H_2=\{ab\colon a\in H_1,\; b\in H_2\}\) is partitioned as

\(\displaystyle H_1H_2=\left(\bigsqcup_{i=1}^{n-1} h_iK\right)H_2=\bigsqcup_{i=1}^{n-1} h_iKH_2=\bigsqcup_{i=1}^{n-1} h_iH_2. \)

The last equality holds because \(\displaystyle K\subseteq H_2\), so \(\displaystyle KH_2=H_2\). The last expression is a disjoint union since

\(\displaystyle h_iH_2\cap h_jH_2\neq\varnothing\iff h_i^{-1}h_j\in H_2\iff h_i^{-1}h_j\in K\iff h_i=h_j. \)

Thus, \(\displaystyle H_1H_2\) is a disjoint union of \(\displaystyle n-1\) left cosets with respect to \(\displaystyle H_2\); hence \(\displaystyle L=G\setminus(H_1H_2)\) is the remaining such left coset. Similarly, \(\displaystyle L\) is a right coset with respect to \(\displaystyle H_1\). Therefore, for each \(\displaystyle g\in L\) we have \(\displaystyle L=gH_2=H_1g\), which yields \(\displaystyle H_2=g^{-1}H_1g\).

Solution 2. Put \(\displaystyle G/H_1 = X\) and \(\displaystyle G/H_2 = Y\), those are \(\displaystyle n\)-element sets acted on by \(\displaystyle G\) from the left. Let \(\displaystyle G\) act on \(\displaystyle X\times Y\) from the left coordinate-wise, consider this product as a table, with rows being copies of \(\displaystyle X\) and columns being copies of \(\displaystyle Y\).

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

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

\(\displaystyle n(n-1)=|Z|=|X\times Y|=n^2,\)

which is a contradiction.

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

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

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

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