Galois Theory, 4th Edition

Speculations about Proofs 305

H has q conjugates, which intersect pairwise in the identity, and K has p conjugates,

which intersect pairwise in the identity. Therefore G has 1 element of order 1, at least

(p −1)q elements of order p, and at least p(q −1) elements of order q. These total

2pq − p −q + 1 = pq + (p −1)(q −1) elements, a contradiction since p, q > 1.

(3) Suppose G is simple of order 2p

k

. There is no subgroup of index 2, so every

proper subgroup has index divisible by p, contrary to Lemma 25.4(3).

(4) Suppose G is simple of order 3p

k

. Since 3p

k

≥ 8, Lemma 25.5 implies that there

is no subgroup of index ≤3. Therefore every proper subgroup