Exercise (§1.7, p.53)

Basic Algebra (Solutions)

by Huah Chu

Exercise (§1.7, p.53)

1. Determine the cosets of hαi in S − 4 where α = (1234).

Ans. Let H = h(1234)i. The right cosets are H, H(12) = {(12), (134), (1423), (243)}, H(13) = {(13), (14)(23), (24), (12)(34)}, H(14) = {(14), (234), (1243), (132)}, H(23) = {(23), (241), (1342), (143)}, H(24) = {(24), (12)(34), (13), (14)(23)}. The left coset of H are left for the reader.

2. Show that if G is finite and H and K are subgroups such that H ⊃ K then [G : K] = [G : H][H : K].

Proof. (I) Since H is a subgroup of G, hence |G| = [G : H]|H|. K is a sub- group of H, |H| = |K|[H : K]. Thus |G| = [G : H][H : K]|K|, K is a sub- group of G, |G| = [G : K]|K|. Hence [G : H][H : K]|K| = [G : H]|K| and [G : H][H : K] = [G : K].

(II) Let G = S_n

i=1Hh_i, H = S_m

j=1Kk_j where n = [G : H], m = [H : K]. Then G = S

1≤i≤n

1≤j≤mKk_jh_i. It is easy to check that Kk_jh_i 6= Kk_rh_s if (j, i) 6= (r, s). Hence

[G : H] = nm = [G : H][H : K]. ¤

3. Let H₁ and H₂ be subgroups of G. Show that any right coset relative to H₁ ∩ H₂ is the intersection of a right coset of H₁ with a right coset of H₂. Use this to prove Poincare’s Theorem that if H₁ and H₂ have finite index in G then so has H₁∩ H₂. Proof. (1) Let (H₁ ∩ H₂)x be any coset of H₁ ∩ H₂, we just need to prove that (H1∩ H2)x = H1x ∩ H2x:

For y ∈ H₁x ∩ H₂x, y = h₁x for h₁ ∈ H₁. Since h₁x ∈ H₂x, h₁ = (h₁x)x⁻¹ ∈ H₂, so h₁ ∈ H₁∩ H₂. y ∈ (H₁∩ H₂)x.

(2) Let {H₁x₁, . . . , H₁x_n} be cosets of H₁ and {H₂y₁, . . . , H₂y_m} cosets of H₂. From (1) any cosets (H₁∩ H₂)x of H₁∩ H₂ is the intersection of a right coset H₁x_i of H₁ with a right coset H2yj of H2. Hence H1∩ H2 has only a finite number (≤ nm) of cosets. ¤ 4. Let G be a finitely generated group, H a subgroup of finite index. Show that H is finitely generated.

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Proof. Let S = {g1, . . . , gm} be a finite generating set of G. We may assume that g_i⁻¹ ∈ S for all i. Let {Hx₁, Hx₂, . . . , Hx_n} be the right cosets of H, where x₁ = 1.

For any i, j, x_ig_j = u_ijx_i, for some u_ij ∈ H and some coset representative x_i⁰. We shall show that H is generated by {uij}, hence is finitely generated.

Let h = g_i1g_i2· · · g_il ∈ H, where g_ij ∈ H. Then

h = (x₁g_i1)g_i2· · · g_il = (u_1i1x₁⁰)g_i2· · · g_il

= u_1i1(x₁⁰g_i2)g_i3· · · g_il = u_1i1(u_j⁰_ilx₂⁰)g_i3· · · g_il

= · · ·

= u_1i1u₁⁰_i2· · · u_(l−1)⁰_ilx_s⁰ ∈ H = Hx₁. Hence x_s⁰ = x₁ = 1 and H is generated by {u_ij}.

Remark. If H is any subgroup of a finitely generated group G, it is not necessary that H should be finitely generated. In fact, the commutator subgroup of a free group of rank two is not finitely generated.

5. Let H and K be two subgroups of a group G. Show that the set of maps x → hxk, h ∈ H, k ∈ K is a group of transformations of the set G. Show that the orbit of x relative to this group is the set HxK = {hxk|h ∈ H, k ∈ K}. This is called the double coset of x relative to the pair (H, K). Show that if G is finite then |HxK| = |H|[K : x⁻¹Hx ∩ K].

Proof. We only prove the last statement. We write M = x⁻¹Hx ∩ K for simplicity.

We shall show that the mapping Mk → Hxk establishes a one to one correspondence between the cosets of x⁻¹Hx∩K in K and the cosets of H is HxK. Thus |HxK|/|K| = [K : x⁻¹Hx ∩ K], hence the result.

(i) The mapping is well-defined. If Mk = Mk⁰, then k(k⁰)⁻¹ ∈ M = x⁻¹Hx ∩ K, k(k⁰)⁻¹ ∈ x⁻¹Hx, xkk⁰⁻¹x⁻¹ = xk(xk⁰)⁻¹ ∈ H. Thus Hxk = Hxk⁰.

(ii) The mapping is one to one. Reversing the implications in (i) will get (ii).

(iii) The mapping is onto obviously. ¤

Remark. Let H and K be subgroups of G. Then HxK is an orbit under the trans- formation group stated in the exercise. Hence G has a double coset decomposition G =S

x∈GHxK.

6. Let H be a subgroup of finite index in a group G. Show that there exists a set of elements z₁, z₂, . . . , z_r ∈ G, r = [G : H], which are representatives of both the set of right and the set of left cosets, that is, G is the disjoint union of the Hz_i and also of the ziH.

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Proof. Let H be a subgroup of G. By the remark after exercise 5, G has a double coset decomposition G = Hg₁H ∪ Hg₂H ∪ · · · ∪ Hg_lH. To prove this exercise, it is enough to show that, for each double coset Hg_iH, there exist z₁, . . . , z_s so that they are representatives of the left and the right cosets of H contained in the double coset Hg_iH.

Let HgH be any double coset. Write HgH = S_s

i=1Hgx_i where x_i ∈ H and Hgxi ∩ Hgxj = ∅ if i 6= j. We also write HgH = S_s⁰

i=1yigH where yi ∈ H and y_igH ∩ y_igH = ∅ if i 6= j. ¿From exercise 5, we have s = s⁰. For any x_i, y_i ∈ H we have Hgx_i = Hy_igx_i and y_igx_iH = y_igH. Hence {y_igx_i} are the representatives which

we want. ¤

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