Let \(G\) be a finite group and \(H\) be a subgroup of \(G\) of index \(n\). Show that \(H\) has a subgroup \(K\) such that \(K\) is normal in \(G\) and \([G : K]\) divides \(n!\).
Solution
Take the core of \(H\) in \(G\), \[ K=\bigcap_{x\in G}xHx^{-1}. \]
1. Left multiplication defines an action of \(G\) on theset \(X=\{xH:x\in G\}\) of its \(n\) left cosets.
For \(g\in G\), set \(\rho(g)(xH)=gxH\). If \(xH=yH\), then \(x=yh\) for some \(h\in H\), and \(gxH=gyhH=gyH\). Thus the map is independent of the chosen representative. Its inverse is \(\rho(g^{-1})\), and \(\rho(gh)(xH)=ghxH=\rho(g)\rho(h)(xH)\). Consequently \(\rho:G\to\operatorname{Sym}(X)\) is a group homomorphism.
2. Its kernel is precisely \(K\), which is normal in \(G\)and contained in \(H\).
An element \(g\) fixes the coset \(xH\) exactly when \(gxH=xH\), or equivalently \(x^{-1}gx\in H\). It fixes every coset exactly when \(g\in xHx^{-1}\) for every \(x\in G\). This proves \(\ker\rho=K\). A kernel is a normal subgroup, and the term \(x=1\) in the intersection shows \(K\subseteq H\).
3. The index \([G:K]\) divides \(n!\).
The first isomorphism theorem gives \(G/K\cong\rho(G)\leq\operatorname{Sym}(X)\) [@DF04]. There are \(n!\) permutations of a set of \(n\) elements. Lagrange’s theorem therefore implies \[ [G:K]=|\rho(G)|\mid|\operatorname{Sym}(X)|=n!. \] If \(n=1\), the action is trivial and the construction gives \(K=H=G\), with index \(1=1!\).