Let \(G\) be a finite group of order \(n\) and let \(p\) be the smallest prime dividing \(n\). Show that any subgroup of \(G\) of index \(p\) is normal in \(G\).
Solution
Let \(H\leq G\) have index \(p\).
1. Left multiplication on the \(p\) left cosets of \(H\) gives ahomomorphism \(\rho:G\to S_p\), and the stabilizer of the coset \(H\)is exactly \(H\).
For \(g\in G\), define \(\rho(g)(aH)=gaH\). Equality of left cosets is preserved by left multiplication, so the map is well-defined. It is a permutation with inverse \(\rho(g^{-1})\), and associativity gives \(\rho(gk)=\rho(g)\rho(k)\). Finally, \(\rho(g)(H)=H\) if and only if \(gH=H\), which is equivalent to \(g\in H\). In particular, \(\ker\rho\subseteq H\).
2. The subgroup \(H\) is the kernel of \(\rho\), and is therefore normal.
Every permutation in \(\rho(H)\) fixes the coset \(H\), so it acts as a permutation of the other \(p-1\) cosets. Thus \(\rho(H)\) is a subgroup of \(S_{p-1}\), and its order divides \((p-1)!\). Its order also divides \(|H|\), by the first isomorphism theorem and Lagrange’s theorem [@DF04].
Every prime dividing \(|H|\) divides \(|G|=n\), so is at least \(p\). Every prime dividing \((p-1)!\) is less than \(p\). Hence these two integers are relatively prime. This includes \(p=2\), when \((p-1)!=1\). The two divisibilities imply \(|\rho(H)|=1\). Consequently \(H\subseteq\ker\rho\), and step 1 gives the reverse containment. Therefore \(H=\ker\rho\lhd G\).