Bourbaki

Exercise 20, § 7

Algebra, I, 164

Bourbaki marks this one as harder than the rest.

We preserve the notation of the preceding exercise. Let H be a subgroup of G.

(a) Let RHR_H be the set of S-connected subsets T of G such that hT∩T=∅hT \cap T = \varnothing if h∈H,h≠eh \in H, h \neq e. Show that RHR_H is inductive for the relation of inclusion.

(b) Let Y be a maximal element of RHR_H. Show that G=⋃h∈HhYG = \bigcup_{h \in H} hY, in other words Y is a system of representatives of the right cosets modulo H. (If G′=⋃hYG' = \bigcup hY is distinct from G, show that there exist x∈G′x \in G' and y∈Yy \in Y which are S-neighbours and deduce that Y∪{x}Y \cup \{x\} belongs to RHR_H.)

(c) Let SHS_H be the set of elements h∈Hh \in H such that Y and hYhY are neighbours. It is a symmetric subset of H. Show that an element h∈Hh \in H belongs to SHS_H if and only if h≠eh \neq e and there exist y,y′∈Y,s∈Sy, y' \in Y, s \in S with ys=hy′ys = hy'. Show that SHS_H generates H. (If H′H' is the subgroup of H generated by SHS_H, prove that ⋃h∈H′hY\bigcup_{h \in H'} hY is a connected S-component of G.) Obtain this result by means of Exercise 14 of § 4.

(d) Let XHX_H be a subset of SHS_H such that SHS_H is the disjoint union of XHX_H and XH−1X_H^{-1} (show that such a subset exists).

Show that XHX_H is a basic family for H and in particular that H is a free group ("Nielsen-Schreier Theorem"). (Apply the criterion of Exercise 18(c) to H together with SHS_H, noting that two elements h,h′h, h' of H are SHS_H-neighbours if and only if hYhY and h′Yh'Y are neighbouring subsets of G. Use Exercise 19(d) to prove that SHS_H satisfies condition (ii) in Exercise 18(c).)

(e) Let d=(G:H)=Card(Y)d = (G:H) = \mathrm{Card}(Y). Suppose dd is finite. Show that:

Card(XH)=Card(X)=Card(G)if Card(X) is infiniteCard(XH)=1+d(Card(X)−1)if Card(X) is finite. \begin{aligned} \mathrm{Card}(X_H) &= \mathrm{Card}(X) = \mathrm{Card}(G) & \text{if } \mathrm{Card}(X) \text{ is infinite} \\ \mathrm{Card}(X_H) &= 1 + d(\mathrm{Card}(X) - 1) & \text{if } \mathrm{Card}(X) \text{ is finite}. \end{aligned}

(Let T be a non-empty S-connected finite subset of G and T' (resp. T'') the set of ordered pairs of S-neighbouring elements whose first element (resp. both of whose elements) belongs (resp. belong) to T. Card(T′)=2Card(X)Card(T)\mathrm{Card}(T') = 2\mathrm{Card}(X)\mathrm{Card}(T) and it should be shown by induction on Card(T)\mathrm{Card}(T) that

Card(T′′)=2Card(T)−2. \mathrm{Card}(T'') = 2\mathrm{Card}(T) - 2.

Apply the result to T=YT = Y, noting that SHS_H is equipotent to T′−T′′T' - T''.)

Solution

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Transcribed from the printed volume. Not translated.