Exercise 23, § 5
Algebra, I, 144
Bourbaki marks this one as harder than the rest.
(a) Let be an element of order 2 in the group which is a product of transpositions with mutually disjoint supports. Show that the centralizer of in is isomorphic to , where admits a normal subgroup of order such that is isomorphic to (observe that every element of permutes the fixed points of and also the supports of ).
(b) Show that, if or , every element of order 2 in whose centralizer is isomorphic to that of a transposition is a transposition (compare the Jordan-Hölder quotients of these centralizers using Exercise 16). Deduce that every automorphism of permutes the transpositions.
(c) If are distinct elements of , let (resp. ) denote the stabilizer of (resp. the fixer of ) in and the transposition . Let be distinct elements of . Show that, if , the group generated by and is (show that this group contains all the transpositions; for this use the fact that, if , then ). Show that the group generated by and is (use the same equation as in the above case).
(d) Show that an automorphism of such that for all is the identity (show that maps every transposition to itself). Show that, if there exists such that , is an inner automorphism (show that there exists an inner automorphism such that maps each of the groups , to itself).
(e) Let be two distinct elements of and let be the centralizer of in . Show that, if , the derived group of is of index 4 in (use the fact that the derived group of is ). Show that is the unique subgroup of of index 2 which does not contain and is not contained in .
(f) Show that, if , an automorphism of which leaves fixed a transposition is inner. (Show by means of (e) that ; show equally that, if , is either of the form , , or of the form , ; in the first case conclude that and apply (d); reduce the second case to the first by multiplying by the inner automorphism of defined by .)
(g) Deduce from (b) and (f) that every automorphism of is inner if or .
Solution
No solution has been written yet.