Exercise 14, § 10
Algebra, II, 421
Bourbaki marks this one as harder than the rest.
Let E be a vector space of dimension over a field K.
(a) Show that the projective group is canonically isomorphic to the quotient of the linear group by its centre (isomorphic to the multiplicative group of the centre of K, cf. § 1, Exercise 26).
(b) Let denote the canonical image of the group in ; it is a normal subgroup of , containing the commutator group of when is finite-dimensional. Show that is a doubly transitive (cf. I, § 5, Exercise 14) group of permutations of .
(c) Let be an element of and let (in the notation of § 9, no. 6). Show that the transvections of the form (where is such that ) constitute a commutative subgroup of ; if is the image of this subgroup in , is a normal subgroup of the subgroup of leaving invariant. Show also that the union of the subgroups conjugate to in generates .
(d) Let be a primitive group (I, § 5, Exercise 13) of permutations of a set which is equal to its commutator group. Suppose that there exists an element such that the subgroup of leaving invariant contains a commutative normal subgroup such that the union of the subgroups conjugate to in generates . Show that under these conditions is simple. (Let be a normal subgroup of distinct from the identity; using I, § 5, Exercise 17, show that and, using the fact that is commutative, deduce that every commutator in is contained in .)
(e) Deduce from (c), (d) and Exercise 13(c) that the group is simple except when and has 2 or 3 elements.
(f) If and is a field with elements, show that the order of is
(g) Show that if and is a field with 2 (resp. 3) elements, is isomorphic to the symmetric group (resp. the alternating group ). (Considering as a group of permutations of , note that it contains every transposition (resp. every cycle and use f).)
(h) Show that, except in the cases considered in (g), every normal subgroup of distinct from is contained in the centre of (use (d) and the fact that every transvection is a commutator in (Exercise 13(c))).
Solution
No solution has been written yet.