Bourbaki

Exercise 24, § 5

Algebra, I, 145

(a) Show that there exists a subgroup HH of S6\mathfrak{S}_6 which is isomorphic to S5\mathfrak{S}_5 and leaves fixed no element of {1,6}\{1, 6\} (let S5\mathfrak{S}_5 operate by inner automorphisms on the set of its subgroups of order 5, which has 6 elements).

(b) Show that there exists an automorphism σ\sigma of S6\mathfrak{S}_6 which maps (1 2)(1\ 2) to (1 2)(3 4)(5 6)(1\ 2)(3\ 4)(5\ 6). (Let S6\mathfrak{S}_6 operate on S6/H\mathfrak{S}_6/H, where HH is chosen as above.)

(c) Show that the group of inner automorphisms of S6\mathfrak{S}_6 is of index 2 in the group of all automorphism of S6\mathfrak{S}_6. (Let α\alpha be an automorphism of S6\mathfrak{S}_6; show that α\alpha maps a transposition either to a transposition or to a conjugate of (1 2)(3 4)(5 6)(1\ 2)(3\ 4)(5\ 6). Using Exercise 23, show that in the first (resp. second) case α\alpha (resp. α∘σ\alpha \circ \sigma) is an inner automorphism.)

Solution

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Transcribed from the printed volume. Not translated.