Let \(G\) be a finite group of order \(p^k m\) where \(p\) is prime and \((p,m) = 1\). Let \(n_p\) be the number of subgroups of \(G\) of order \(p^k\). State as many facts as you can about the value of \(n_p\). (No proofs are required for your answers in this part.)
Show that a non-abelian simple group \(G\) has no proper subgroup of index \(\leq 4\).
(You may assume without proof that the alternating group \(A_5\) is the smallest non-abelian simple group.)
Solution
1. For part (a), the Sylow count satisfies\[ 1\leq n_p\leq m,\qquad n_p\mid m,\qquad n_p\equiv1\pmod p. \]For any Sylow \(p\)-subgroup \(P\), one also has\[ n_p=[G:N_G(P)]. \]Moreover, \(n_p=1\) if and only if \(G\) has a normal Sylow\(p\)-subgroup. In that case this Sylow subgroup ischaracteristic. If \(n_p>1\), then \(n_p\geq p+1\).
Existence, conjugacy, divisibility, and the congruence are the Sylow theorems [@DF04]. The bound \(n_p\leq m\) follows from the positive divisibility \(n_p\mid m\). Conjugation acts transitively on the Sylow subgroups, and the stabilizer of \(P\) is its normalizer \(N_G(P)\); the orbit-stabilizer formula gives the displayed index.
A unique Sylow subgroup is fixed by all conjugations, so is normal. Conversely, normality makes its conjugacy orbit a singleton, and Sylow conjugacy then gives uniqueness. Every automorphism of \(G\) preserves subgroup orders and therefore permutes the Sylow \(p\)-subgroups; a unique one is consequently characteristic. Finally, a positive integer larger than \(1\) and congruent to \(1\) modulo \(p\) is at least \(p+1\).
2. A nonabelian simple group has no proper subgroupof index at most \(4\).
Suppose \(H<G\) has finite index \(n\leq4\). Properness implies \(2\leq n\leq4\). The action on the left cosets defines a homomorphism \[ \rho:G\longrightarrow\operatorname{Sym}(G/H)\cong S_n, \qquad \rho(g)(aH)=gaH. \] Its kernel is a normal subgroup of \(G\). It is not all of \(G\): if \(g\notin H\), then \(\rho(g)\) moves the coset \(H\) to the different coset \(gH\). Simplicity therefore makes the kernel trivial.
Thus \(G\) embeds in \(S_n\), so it is finite and \(|G|\leq n!\leq24\). Since \(A_5\) is the smallest nonabelian simple group, \(|G|\geq|A_5|=5!/2=60\), a contradiction.