Exercises (§1.11, p.69)

Basic Algebra (Solutions)

by Huah Chu

Exercises (§1.11, p.69)

1. Let S be a subset of a group G such that g⁻¹Sg ⊂ S for any g ∈ G. Show that the subgroup hSi generated by S is normal. Let T be any subset of G and let S =S

g∈Gg⁻¹T g. Show that hSi is the normal subgroup generated by T .

Proof. Omitted. ¤

The following three exercises are taken from Burnside’s The Theory of Groups of Finite Order, 2nd ed., 1911. (Dover reprint, pp.464–465.)

2. Using the generators (12), (13), . . . , (1n) (See exercise 5, §1.6) for S_n, show that S_n is defined by the following relations on x₁, x₂, . . . , x_n−1 in F G⁽ⁿ⁻¹⁾:

x²_i, (x_ix_j)³, (x_ix_jx_ix_k)², i, j, k 6= .

Proof. (I) Step 1. In S_n, let (1i) = x_i−1, then they satisfy x²_i = 1, (x_ix_j)³ = (1ji)³ = 1, (xixjxixk)² = (ij)²(1k)² = 1, for i, j, k, distinct. Hence Snis a homomorphic image of F G⁽ⁿ⁻¹⁾/K where K is the normal subgroup generated by the above relations.

Step 2. We shall now show that |F G⁽ⁿ⁻¹⁾/K| ≤ n! which will imply that S_n ' F G⁽ⁿ⁻¹⁾/K. Let H be the subgroup generated by x1, x2, . . . , xn−2. It is enough to prove that F G⁽ⁿ⁻¹⁾/K : H] ≤ n and then get our result by induction.

Step 3. We prove that for any g, g can be written in the form of h, hx_n−1 or hx_n−1x_i for some h ∈ H, 1 ≤ i ≤ n − 2.

Note that the relation (x_ix_j)³ = 1 will imply

(α) x_n−1x_ix_n−1= x_ix_n−1x_i, 1 ≤ i ≤ n − 2. The relation (x_ix_jx_ix_k)² = 1 will imply (β) xn−1xixj = xixjxixn−1xi, i, j, n − 1 are distinct. Given any g ∈ F G⁽ⁿ⁻¹⁾/K write g as a word in x₁, . . . , x_n−1, say, g = x_i1· · · x_ilx_n−1x_j1· · · x_jm where x_i1· · · x_il ∈ H.

We write x_i1· · · x_il = h for simplicity. We also assume x_j1 6= x_n−1 since x²_n−1 = 1. We prove the assertion by induction on m.

(1) If x_j2 = x_n−1, then g = hx_j1x_n−1x_j1x_j3· · · x_jm by (α),

(2) If x_j2 6= x_n−1, then g = hx_j1x_j2x_j1x_n−1x_j1x_j3· · · x_jm by (β). In any case, g = h⁰x_n−1x_j1x_j3· · · x_jm for h⁰ ∈ H. The number m is reduced. By induction, it is easy to see that g can be transformed into the desired forms. ¤

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Proof (II). Induction on n, n = 2, 3. Now let

G = hx₁, . . . , x_n−1i, H = hx₁, . . . , x_n−2i.

Define a homomorphism φ

φ : G → S_n

xi 7→ (1, i + 1).

φ is an epimorphism. To prove that φ is an isomorphism, it suffices to show that

|G| ≤ n⁰_·. ¤

By induction hypothesis, H ' S_n−1. Consider eH ^def= H ∪ Hx_n−1∪ Hx_n−1x₁x_n−1∪

· · · ∪ Hx_n−1x_n−2x_n−1. If we can show that eH = G, then |G| ≤ n · |H| = n⁰_·. We claim that eHx_i ⊂ eH for all i = 1, 2, . . . , n − 1.

Assuming the above claim, we find that xi ∈ eH (since 1 ∈ H ⊂ eH) and eH is closed under multiplication. It follows that eH = G.

Now we shall prove that eHx_i ⊂ eH.

Case 1. 1 ≤ i ≤ n − 2.

H · xi = H ⊂ eH

Hx_n−1· x_i = H(x_n−1x_i) = H(x_ix_n−1)² = Hx_i· x_n−1x_ix_n−1 = Hx_n−1x_ix_n−1⊂ eH.

Hx_n−1x_jx_n−1· x_i =

½Hx_ix_n−1 = Hx_n−1 ⊂ eH if j = i.

Hx_ix_n−1x_jx_n−1 = Hx_n−1x_jx_n−1⊂ eH if j 6= i.

Case 2. i = n − 1.

H · xn−1 = Hxn−1 ⊂ eH Hxn−1· xn−1= H ⊂ eH

Hxn−1xjxn−1· xn−1 = Hxn−1xj = Hxjxn−1xjxn−1= Hxn−1xjxn−1 ⊂ eH.

3. Using the generators (12), (23), . . . , (n − 1n) for S_n show that this group is defined by x₁, . . . , x_n−1 subjected to the relations:

x²_i, (xixi+1)³, (xixj)², j > i + 1.

Proof. Step 1. Let x_i = (i, i + 1), it is easy to see that they satisfy the relations: x²_i, (xixi+1)³, (xixj)², j > i + 1.

Step 2. By the similar arguments as in exercise 2, it is enough to show that F G⁽ⁿ⁻¹⁾/K = H ∪Hx_n−1∪Hx_n−1x_n−2∪· · ·∪Hx_n−1x_n−2· · · x₁ where H is the subgroup generated by x1, . . . , xn−2:

Let g = hx_n−1x_j1· · · x_jm ∈ F G⁽ⁿ⁻¹⁾/K where h ∈ H, we prove this assertion by induction on m. We first note that the relations (x_ix_i+1)³ = 1 and (x_ix_j)² = 1, j > i+1 imply that

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(α) xjxi = xixj, if j > i + 1, (β) x_n−1x_n−2x_n−1= x_n−2x_n−1x_n−2.

(i) If x_j1 6= x_n−2, then g = hx_j1x_n−1x_j2· · · x_jm by (α).

(ii) If xj1 = xn−2 and xj2 6= xn−1 (xj2 6= xn−2), then g = hxn−1xn−2xj2· · · xjm = hx_n−1x_j2x_n−2x_j3· · · x_jm (by (α)) = hx_j2x_n−1x_n−2x_j3· · · x_jm (by (α)).

(iii) If x_j1 = x_n−2, x_j2 = x_n−1, then g = hx_n−1x_n−2x_n−1x_j3· · · x_jm = hx_n−2x_n−1x_n−2x_j3· · · x_jm by (β).

In any case, the number m is reduced. And the proof is completed.

Remark. As in proof (II) of the above exercise, we can show that eHx_i ⊂ eH where He ^def= H ∪ Hxn−1∪ Hxn−1xn−2∪ · · · ∪ Hxn−1xn−2· · · x1.

4. Show that A_n can be defined by the following relations on x₁, x₂, . . . , x_n−2 x³₁; x²_i, i > 1; (xixi+1)³; (xixj)², j > i + 1.

Proof. Step 1. In A_n, set x₁ = (123), x_i = (12)(i + 1, i + 2) for 2 ≤ i ≤ n − 2. It is easy to verify that they satisfy the given relations.

Step 2. Similar to the arguments in exercise 2, it is enough to show that [F G⁽ⁿ⁻²⁾/K : H] ≤ n where H is the subgroup generated by x₁, x₂, . . . , x_n−3. For this purpose, we shall show that F G⁽ⁿ⁻²⁾/K = H ∪ Hx_n−2 ∪ Hx_n−2x_n−3∪ Hx_n−2x_n−3x_n−4∪ · · · ∪ Hxn−2xn−3· · · x2x1∪ Hxn−2· · · x2x²₁:

For any g = hx_n−2x_j1· · · x_jm ∈ G where h ∈ H, we shall prove this assertion by induction on m. ¿From the given relations we have (α) x_jx_i = x_ix_j, j > i + 1; and (β) xn−3xn−2xn−3 = xn−2xn−3xn−2.

The remaining proof are quite similar to exercise 3 and left to the reader. ¤

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