State Sylow’s Theorem.
Prove that no group of order 6545 is simple.
Solution
1. Sylow’s theorem has the following form [@DF04]. If \(H\) is finiteand \(|H|=p^a m\) with \(p\) prime and \(p\nmid m\), then:
- There is a subgroup of order \(p^a\), called a Sylow \(p\)-subgroup.
- Every \(p\)-subgroup lies in a Sylow \(p\)-subgroup, and all Sylow \(p\)-subgroups are conjugate.
- Their number \(n_p\) satisfies \(n_p\mid m\) and \(n_p\equiv1\pmod p\). For any Sylow \(p\)-subgroup \(P\), it is also \(n_p=[H:N_H(P)]\).
2. If \(|G|=6545\), the number of Sylow \(5\)-subgroups is \(1\) or \(11\).
The factorization is \(6545=5\cdot7\cdot11\cdot17\). Hence Sylow’s theorem requires \(n_5\mid1309=7\cdot11\cdot17\) and \(n_5\equiv1\pmod5\). The positive divisors of \(1309\) are \[ 1,\ 7,\ 11,\ 17,\ 77,\ 119,\ 187,\ 1309. \] Among these, only \(1\) and \(11\) are congruent to \(1\) modulo \(5\).
3. In either case, \(G\) has a nontrivial proper normal subgroup.
If \(n_5=1\), the unique Sylow \(5\)-subgroup is normal, and its order \(5\) makes it nontrivial and proper.
If \(n_5=11\), conjugation on the eleven Sylow \(5\)-subgroups defines \(\rho:G\to S_{11}\). The action is transitive by Sylow conjugacy and has more than one point, so \(\rho\) is nontrivial and \(\ker\rho\ne G\). It cannot be injective: otherwise Lagrange’s theorem would give \(6545\mid11!\), whereas \(17\mid6545\) and \(17\nmid11!\). Thus \(1\ne\ker\rho\ne G\), and the kernel is normal. In both cases \(G\) is not simple.