Let \(p\) be the smallest prime dividing the order of the finite group \(G\). Show that any subgroup of \(G\) of index \(p\) is normal in \(G\).
Solution
Let \(H\leq G\) have index \(p\).
1. There is a normal subgroup \(K\lhd G\) with \(K\leq H\)and \([G:K]\mid p!\).
Act on the set \(X=G/H\) of left cosets by left multiplication. The rule \(\rho(g)(xH)=gxH\) is well-defined, since replacing \(x\) by \(xh\), with \(h\in H\), does not change \(gxH\). Its inverse is \(\rho(g^{-1})\), and \(\rho(g_1g_2)=\rho(g_1)\rho(g_2)\), so it defines a homomorphism \(\rho:G\to\operatorname{Sym}(X)\).
Its kernel \(K\) is normal. Every \(k\in K\) fixes \(H\in X\), so \(kH=H\) and \(k\in H\). The first isomorphism theorem identifies \(G/K\) with \(\rho(G)\) [@DF04]. Since \(|X|=p\), the group \(\operatorname{Sym}(X)\) has order \(p!\). Lagrange’s theorem therefore gives \([G:K]\mid p!\).
2. The subgroup \(K\) equals \(H\).
Put \(m=[H:K]\). The index formula gives \[ [G:K]=[G:H][H:K]=pm. \] Since \(pm\mid p!=p(p-1)!\), cancellation of the positive integer \(p\) gives \(m\mid(p-1)!\). Also \(m\) divides \(|H|\), which divides \(|G|\).
If \(m>1\), it has a prime divisor \(q\). Divisibility by \((p-1)!\) implies \(q\leq p-1\), whereas \(q\mid|G|\) and the definition of \(p\) imply \(q\geq p\). This is a contradiction. Hence \(m=1\) and \(H=K\). By step 1, \(H\) is normal in \(G\).