Exercises (§1.4, p.42)

Basic Algebra (Solutions)

by Huah Chu

Exercises (§1.4, p.42)

1. Let A be a monoid, M(A) the monoid of transformations of A into itself, A_L the set of left translations aL, and AR the set of right translations aR. Show that A_L (respectively A_R) is the centralizer of A_R (respectively A_L) in M(A) and that A_L∩ A_R = {c_R= c_L|c ∈ C}, C the center of A.

Proof. (1) We show that A_L = C_{M (A)}(A_R). The case of A_R = C_{M (A)}(A_L) is quite similar.

Since (a_Rb_L)(x) = a_R(bx) = (bx)a = b(xa) = b_L(xa) = (b_La_R)(x), hence A_L ⊆ C_{M (A)}(A_R). Given any ρ ∈ C_{M (A)}(A_R). For all a ∈ M, a_Rρ = ρa_R. In particular, a_Rρ(1) = ρa_R(1), ρ(1)a = ρ(1 · a) = ρ(a). This means that ρ = (ρ(1))_L ∈ A_L.

(2) A_L∧ A_R= {c_R= c_L|c ∈ C}:

It is clearly that c_R = c_L for c ∈ C. Given any ρ ∈ A_L∧ A_R, ρ = a_L for some a.

Since ρ ∈ A_R = C(A_L), hence, for all b ∈ M, a_Lb_L(1) = b_La_L(1). Thus ab = ba and

a ∈ C. So ρ ∈ {c_L|c ∈ C}. ¤

2. Show that if n ≥ 3, then the center of S_n is of order 1.

Proof. Given any 1 6= α ∈ S_n, there exists i such that α(i) 6= 1, say α(i) = j. Choose k 6= i, j since n ≥ 3. Take γ be any permutation in S_n such that γ(i) = i and γ(j) = k.

Then γα(i) = γ(j) = k, and αγ(i) = α(i) = j. Hence γα 6= αγ and α /∈ C(S_n). ¤ Remark. For any α ∈ S_n (n ≥ 3), we can find β ∈ S_n such that it has the same cycle decomposition as α and α 6= β (§1.6). Then there exists γ such that γαγ⁻¹ = β (Ex.

4, §1.6) and γα 6= αγ.

3. Show that any group in which every a satisfies a² = 1 is abelian. What if a³ = 1 for every a?

Proof. (1) Note that the condition a² = 1 implies a = a⁻¹ for all a ∈ G. For any a, b ∈ G, since (ab)² = 1 it follows that ab = (ab)⁻¹. But (ab)⁻¹ = b⁻¹a⁻¹= ba. Hence ab = ba.

(2) If a³ = 1 for all a ∈ G, G need not be abelian. We shall use Sylow’s Theorem and group extensions to construct a counterexample.

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Let G be a finite nonabelian group such that a³ = 1 for all a ∈ G. By Sylow’s Theorem (§1.13), |G| = 3ⁿ. If |G| = 9, G is abelian (§1.12, exercise 6). Hence we assume that |G| = 27.

G contains a normal subgroup K of order 9 (exercise 5.2, p.87 in Rotman: The theory of groups, An introduction). K must be elementary abelian, that is, K = ha, b|a³ = b³ = 1, ab = bai. G/K ' H = hhi is a cyclic group of order 3. Then G is an extension of K by H (see the Remark after exercise 9, §1.12).

h induces an automorphism α of K. If we use the additive notation for composition of K, then K can be regard as a 2-dimensional vector space over finite fieldF3 and α can be represented as a matrix in M2(F3). By suitable changing the basis, assume α has the rational form

· 0 1 a b

¸

(§3, 10). Since α³ = 1, it is not difficult to show that the only solution is α =

· 0 1

−1 −1

¸ .

We have known Ext²_Z[H](Z, H) = H²(H, K) and H²(H, K) = K^H/NK for finite cyclic group H, where K^H = {k ∈ K|αk = k}, NK = {(1 + α + α²)k|k ∈ K}. Hence to find all extensions of K by H, it suffices to compute H²(H, K) first. It is easy to find that αk = k ⇒ k = (x, x) for x ∈ F3 and 1 + α + α² = 0. Hence NK = 0 and H²(H, K) =Z/3Z.

We first check the trivial case, that is, semi-direct product of H by K. Since

· 0 1 1 −1

¸ · 0 1

−1 −1

¸ · 0 1 1 −1

¸₋₁

=

· 1 1 0 1

¸

, we change α to

· 1 1 0 1

¸

for simplic- ity. Then G = ha, b, c|a³ = b³ = c³ = 1, ab = ba, ac = ca, bc = cabi.

Now we check that x³ = 1 for all x ∈ G:

First note that a is in the center of G and from bc = cab, we have cb = a²bc, bcb² = ac and c²bc = ba. Thus

(bc)³ = b(cb)cbc = ba²b(ccbc) = ba²bba = a³b³ = 1.

The verification of (bc²)³ = (b²c)³ = (b²c²)³ = 1 are similar, and left to the reader.

Moreover, since (ax)ⁿ= aⁿxⁿ, so (aⁱb^jc^k)³ = a³ⁱ(b^jc^k)³ = 1. Hence the result. Thus G

is the desired counter-example. ¤

Remarks: (1) Since H²(H, K) ' Z/3Z, we can find a nontrivial factor set f : H × H → K defined by f (h, h) = ab, f (h², h) = f (h, h²) = 1, f (h², h²) = a²b², and f (1, x) = f (x, 1) = 1 for x ∈ H. Let G be the set of all pairs (k, x) ∈ K × H with the composition

(k, x)(k⁰, y) = (k(xk⁰)f (x, y), xy).

Where xk⁰ is defined as follows: (i) If x = 1 then xk⁰ = k⁰; (ii) x = h, k⁰ = aⁱb^j then xk⁰ = a^i+jb^j; (iii) if x = h², k⁰ = aⁱb^j, then xk⁰ = a^i+2jb^j. Then

(1, h)³ = (1 · (h1) · ab, h²)(1, h) = (ab, h²)(1, h) = (ab, 1).

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Thus (1, h) has order 9, which does not satisfy our condition.

(2) The above discussions also hold for any odd prime p. That is, G = ha, b, c|a^p = b^p = c^p = 1, ab = ba, ca = ac, bc = cabi is nonabelian group such that x^p = 1 for all x ∈ G.

(3) For the group extensions and cohomology of groups, we refer to J. J. Rotman:

The theory of groups; An introduction; or S. MacLane: Homology.

(4) This exercise may be regard as a special case of Burnside’s problem: Let G be finitely generated and n is the l.c.m of the orders of elements, is G a finite group? This exercise shows that, if n = 2, G is abelian. In fact, if n = 3, G is a finite nilpotent group of class ≤ 3. If n = 4 or 6, G is a finite group. We refer the reader to B. Huppert:

Endlich Gruppen, or M. Hall: The theory of groups, Chap. 18, for more defails.

4. For a given binary composition define a simple product of the sequence of elements a1, a2, . . . , an inductively as either a1u where u is a simple product of a2, . . . , an or as va_nwhere v is a simple product of a₁, . . . , a_n−1. Show that any product of ≥ 2^relements can be written as a simple product to r elements (which are themselves products).

Proof. We prove it by induction on r. There is nothing to prove for r = 1. Any product of n elements a₁, . . . , a_n, n = 2^r, r > 1, has the form of (a₁, . . . , a_i)(a_i+1, . . . , a_n), where (a₁, . . . , a_i) is a product of a₁, . . . , a_i. Then one of the sequence {a₁, . . . , a_i} and {a_i+1, . . . , a_n} has length ≥ 2^r−1, say {a₁, . . . , a_i}. By the inductive hypothesis, (a₁, . . . , a_i) is a simple product of r − 1 elements and (a₁, . . . , a_i)(a_i+1, . . . , a_n) is a

simple product of r elements. ¤

Remark. The condition of ≥ 2^r may be refined to ≥ 2^r−1+ 1.

The reader may try to give a product of 8 elements which is not a simple product of 4 elements.

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