Using covering spaces, show that an index \(n\) subgroup \(H\) of a group \(G\) has at most \(n\) conjugate subgroups \(gHg^{-1}\) in \(G\). Apply this to show that there exists a normal subgroup \(K \subset G\) of finite index with \(K \subset H\). [For the latter statement, consider the intersection of all the conjugate subgroups \(gHg^{-1}\). This is the maximal normal subgroup of \(G\) contained in \(H\).]
Solution
Let \(H\le G\) have index \(n\). Regard \(G\) as the fundamental group of a connected CW complex and let \[ p:\widetilde X_H\to X \] be the connected covering corresponding to \(H\). Its fiber over the basepoint has \(n\) points and can be identified with the coset set \(G/H\).
1. The conjugates of \(H\) are precisely the stabilizers of points of the transitive \(G\)-set \(G/H\).
For the left action of \(G\) on \(G/H\), the stabilizer of the coset \(gH\) is \[ \operatorname{Stab}(gH) =\{k\in G:kgH=gH\} =gHg^{-1}. \] In covering-space language, changing the chosen lift of the basepoint from the point corresponding to \(H\) to the point corresponding to \(gH\) changes the associated subgroup by conjugation to \(gHg^{-1}\).
2. Hence \(H\) has at most \(n\) distinct conjugate subgroups.
There are exactly \(n\) points in \(G/H\) and therefore at most \(n\) distinct point stabilizers. By 1 these are the conjugates of \(H\).
3. Let\[ K=\bigcap_{g\in G}gHg^{-1}. \]Then \(K\) is normal in \(G\) and \(K\subseteq H\).
Containment in \(H\) follows from the term \(g=1\). For \(a\in G\), \[ aKa^{-1} =\bigcap_{g\in G}agHg^{-1}a^{-1} =\bigcap_{u\in G}uHu^{-1} =K, \] where \(u=ag\) ranges over all of \(G\). Thus \(K\trianglelefteq G\).
4. The subgroup \(K\) has finite index in \(G\).
By 2 only finitely many distinct conjugates of \(H\) occur, say \[ H_1,\dots,H_r, \qquad r\le n. \] Each has index \(n\). The diagonal map \[ G/(H_1\cap\cdots\cap H_r) \longrightarrow \prod_{i=1}^r G/H_i, \qquad gK\longmapsto(gH_1,\dots,gH_r) \] is injective. Therefore \[ [G:K]\le\prod_{i=1}^r[G:H_i]=n^r<\infty. \] Equivalently, \(K\) is the kernel of the monodromy homomorphism \[ G\longrightarrow\operatorname{Sym}(G/H)\cong S_n, \] so in fact \([G:K]\le n!\).
5. Thus there is a finite-index normal subgroup\[ \boxed{K\trianglelefteq G,\qquad K\subseteq H.} \]
Combine 3 and 4.