Let \(G\) be a group and let \(p\) be the smallest prime divisor of \(|G|\). Suppose that \(H\) is a subgroup of \(G\) of index \(p\). Prove that \(H\) is normal.
Solution
1. The action on left cosets gives a homomorphism\(\rho:G\to S_p\) whose kernel \(K\) is contained in \(H\).
There are exactly \(p\) left cosets of \(H\). Left multiplication \(g\cdot(xH)=(gx)H\) is well defined and permutes these cosets. Composition agrees with multiplication in \(G\), giving \(\rho\). The action is transitive, since \(yx^{-1}\) sends \(xH\) to \(yH\). Any element of the kernel fixes the coset \(H\), hence belongs to its stabilizer, which is exactly \(H\). Thus \(K\subseteq H\).
2. The image of \(\rho\) has order exactly \(p\).
Let \(m=|\rho(G)|\). The first isomorphism theorem and Lagrange’s theorem give \(m\mid |G|\) and \(m\mid p!\) [@DF04]. Every prime dividing \(m\) must therefore be at least \(p\), by minimality of \(p\), and at most \(p\), since it divides \(p!\). Thus \(p\) is its only possible prime divisor. The exponent of \(p\) in \(p!\) is one, so \(m=1\) or \(m=p\). Transitivity and orbit-stabilizer give \(p\mid m\), hence \(m=p\).
3. The subgroup \(H\) equals \(K\) and is normal.
The index of \(K\) is \([G:K]=|\rho(G)|=p\). Since \(K\subseteq H\), index multiplication gives \[ p=[G:K]=[G:H][H:K]=p[H:K]. \] Therefore \([H:K]=1\) and \(H=K\). A homomorphism kernel is normal, so \(H\lhd G\).