- Prove that every subgroup of index \(2\) is normal.
- More generally, let \(G\) be finite and let \(p\) be the smallest prime dividing \(|G|\). Prove that every subgroup of index \(p\) is normal.
Solution
1. Index \(2\).
If \(H\le G\) has index \(2\), then there are exactly two left cosets and two right cosets. For \(g\notin H\), both \(gH\) and \(Hg\) are the complement \(G\setminus H\). Hence \[ gH=Hg \] for every \(g\in G\), so \(H\trianglelefteq G\).
2. Index equal to the smallest prime divisor.
Let \([G:H]=p\) and let \(G\) act on the left cosets \(G/H\). This gives \[ \rho:G\to S_p. \] The image is transitive, so \(p\mid |\rho(G)|\). Also \(|\rho(G)|\) divides both \(|G|\) and \(p!\).
Every prime divisor of \(|G|\) is at least \(p\), while every prime divisor of \(p!\) is at most \(p\). Therefore the only possible prime divisor of \(|\rho(G)|\) is \(p\). Since \(p^2\nmid p!\), we get \[ |\rho(G)|=p. \] A transitive group of prime order acts regularly, so the stabilizer of the coset \(H\) in \(\rho(G)\) is trivial. But the stabilizer upstairs is \(H\), hence \[ H=\ker\rho. \] Therefore \(H\) is normal in \(G\).