Bourbaki

Exercise 26, § 5

Algebra, I, 146

Bourbaki marks this one as harder than the rest.

(a) Let GG be a group operating on two finite sets E1E_1 and E2E_2. If s∈Gs \in G, let s1s_1 (resp. s2s_2) denote the permutation of E1E_1 (resp. E2E_2) defined by ss and E1sE_1^s (resp. E2sE_2^s) the set of elements invariant under s1s_1 (resp. s2s_2).

Show the equivalence of the following properties: (i) For all s∈Gs \in G, Card(E1s)=Card(E2s)\mathrm{Card}(E_1^s) = \mathrm{Card}(E_2^s). (ii) For all s∈Gs \in G, the orders of the component cycles of s1s_1 (cf. no. 7, Proposition 7) are the same (to within a permutation) as those of s2s_2. (iii) For all s∈Gs \in G, there exists a bijection fs:E1→E2f_s : E_1 \to E_2 such that s2∘fs=fs∘s1s_2 \circ f_s = f_s \circ s_1.

If these properties hold, the GG-sets E1E_1 and E2E_2 are said to be weakly equivalent.

(b) Give an example of two weakly equivalent GG-sets which are not isomorphic (take GG to be a non-cyclic group of order 4).

(c) Let H1H_1 and H2H_2 be two subgroups of a finite group GG. Show the equivalence of the following properties: (1) For every conjugacy class CC of GG,

Card(C∩H1)=Card(C∩H2). \mathrm{Card}(C \cap H_1) = \mathrm{Card}(C \cap H_2).
(2) The GG-sets G/H1G/H_1 and G/H2G/H_2 are weakly equivalent.

(d) Show that, if H1H_1 and H2H_2 satisfy properties (1) and (2) above and H1H_1 is normal in GG, then H2=H1H_2 = H_1.

(e) Let G=A6G = \mathfrak{A}_6 and let

H1={e,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)} H_1 = \{ e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3) \}
H2={e,(1 2)(3 4),(1 2)(5 6),(3 4)(5 6)}. H_2 = \{ e, (1\ 2)(3\ 4), (1\ 2)(5\ 6), (3\ 4)(5\ 6) \}.
Show that H1H_1 and H2H_2 satisfy properties (1) and (2) and that H1H_1 and H2H_2 are not conjugate in A6\mathfrak{A}_6 (nor in S6\mathfrak{S}_6).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.