Exercise 39, § 6
Algebra, I, 158
Bourbaki marks this one as harder than the rest.
Let G be a finite group, , a prime number, X the set of Sylow -subgroups of G and . Then G operates transitively on X by conjugation; let be the corresponding homomorphism (A B denotes the relation "A is isomorphic to B") and K its kernel.
(a) Show that K is the intersection of the normalizers of the and that the Sylow -subgroup of K is the intersection of the . Deduce that if the order of K divides .
(b) Show that if G has no subgroup of index 2, then and that if (resp. ), then (resp. ).
(c) Using (a), (b) and the preceding exercise, show the following facts: If and a Sylow 3-subgroup of G is not normal, then . If and G does not have property (ST), then . If , then G has property (ST). (Show that if a Sylow 3-subgroup of G is not normal, then G contains a subgroup K of order 3 such that and that K is central.) If and G does not have property (ST), then G contains a normal subgroup K of order 2 such that . If and G does not have property (ST), then . (Show that such a group G has no normal subgroup of order 5, nor of order 2, nor of index 2. Deduce that G is isomorphic to a subgroup of and, arguing as in Exercise 25 of § 5, that .)
If and a Sylow 3-subgroup of is not normal, then contains a normal subgroup such that is isomorphic either to or to .
If and a Sylow 2-subgroup of is not normal, then contains a normal subgroup such that .
(d) Show that a group of order is solvable unless .
Solution
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