propiedades grupos

8/4/2019 propiedades grupos

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Arkansas Tech University

MATH 4033: Elementary Modern AlgebraDr. Marcel B. Finan

14 Elementary Properties of Groups

In this section, we prove more theorems about groups. In what follows thegroup binary operation will be referred to as multiplication and thus we willwrite ab.Several simple consequences of the definition of a group are recorded in thefollowing two theorems.

Theorem 14.1

For any group G, the following properties hold:

(i) If a,b,c, G and ab = ac then b = c. (left cancellation law)(ii) If a,b,c, G and ba = ca then b = c. (right cancellation law)(iii) If a G then (a1)1 = a. The inverse of the inverse of an element isthe element itself.(iv) If a, b G then (ab)1 = b1a1. That is the inverse of a product is theproduct of the inverses in reverse order.

Proof.

(i) Suppose that ab = ac. Then

b = eb = (a1a)b= a1(ab) = a1(ac)= (a1a)c = ec = c

(ii) Suppose that ba = ca. Then

b = be = b(aa1)= (ba)a1 = (ca)a1

= c(aa1) = ce = c

(iii) If a G then since aa

1 = a

1a = e then a is an inverse of a

1. Sinceinverses are unique then (a1)1 = a.(iv) Let x be the inverse of ab. Then (ab)x = e By associativity, we havea(bx) = aa1. By (i), we have bx = a1. But bx = ea1 = b(b1a1) so thatby applying (i) again we obtain x = b1a1. Therefore, (ab)1 = b1a1.

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Theorem 14.2

If G is a group and a, b G then each of the equations ax = b and xa = bhas a unique solution. In the first, the solution is x = a1b whereas in thesecond x = ba1.

Proof.

Consider first the equation ax = b. We want to isolate the x on the left side.Indeed, this can be done as follows.

x = ex = (a1a)x= a1(ax) = a1b

To prove uniqueness, suppose that y is another solution to the equation

ax = b. Then, ay = b = ax. By the left cancellation property we have x = y.Finally, the proof is similar for the equation xa = b.

Remark 14.1

Suppose G is a finite group and a, b G. The product ax is in the rowlabelled by a of the Cayley table. By Theorem 14.2, the equation ax = b hasa unique solution means that b appears only once in the row ofa of the table.Thus, each element of a finite group appears exactly once in each row of thetable. Similarly, because there is a unique solution of xa = b, each elementappears exactly once in each column of the Cayley table. It follows that eachrow (respectively column) is a rearrangement of the elements of the group.

Next, we discuss the extension of the associative property to products withany number of factors. More specifically, we will prove the so-called gener-alized associative law which states that in a set with associative operation,a product of factors is unchanged regardless of how parentheses are insertedas long as the factors and their order of appearance in the product are un-changed.

Definition 14.1

For elements a1, a2, , an (n 2) of a group G define

a1a2 an1an = (a1a2 an1)an.

Theorem 14.3 (Generalized Associative Law)For any integer 1 m < n we have

(a1a2 am)(am+1 an) = a1a2 an.

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Proof.

For n 2, let S(n) be the statement(a1a2 am)(am+1 an) = a1a2 an : where a1, a2, , an G and 1 m < n.

We prove that S(n) is true for all n 2 by induction on n. For n = 2the statement is true since (a1)(a2) = a1a2 (the only possible value for m ism = 1.) So assume that the statement is valid for 2, 3, , n 1. We willshow that S(n) is also true. Suppose that 1 m < n. Then either m = n 1or 1 m < n 1. If m = n 1 then

(a1a2 am)(am+1 an) = (a1a2 an1)an = a1a2 an.

So suppose that 1 m < n 1. Then

(a1a2 am)(am+1 an) = (a1a2 am)[(am+1 an1)an]= [(a1a2 am)(am+1 an1)]an= (a1a2 an1)an= a1a2 an

Thus, S(n) is true for all n 2 and this completes a proof of the theorem.

Next, we introduce the concept of integral exponents of elements in a group.The concept plays an important role in the theory of cyclic groups.

Definition 14.2

For any a G we definea0 = ean = an1a, f or n 1

an = (a1)n for n 1.

Remark 14.2

By Theorem 14.3, an = a a a a where the product contains n copies ofa. Also, note that aan1 = an1a = an.

The familiar laws of exponents hold in a group.

Theorem 14.4

Let a be an element of a group G and m and n denote integers. Then

(i) anan = e.(ii) aman = am+n

(iii) (am)n = amn.

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Proof.

(i) The identity is trivial for n = 0. So suppose that n > 0. We use inductionon n. Since aa1 = e then the result is true for n = 1. Suppose the identityholds for 1, 2, , n 1. We must show that it is valid for n. Indeed,

anan = (an1a)[(a1)(n1)a1]= (aan1)[(a1)(n1)a1]= [(aan1)(a1)(n1)]a1

= [a(an1a(n1))]a1

= (ae)a1 = aa1 = e

Thus, the identity is true for all positive integers n.

Now, suppose that n < 0 then

anan = (a1)n(a1)(n) = e.

(ii) We have to show that for a fixed integer m the identity holds for allintegers n. That is, the identity holds for n = 0, n > 0, and n < 0. If m = 0then aman = a0an = ean = an = a0+n for all integers n. So, suppose firstthat m > 0.

n = 0

am+n

=am+0

=am

aman = ama0 = ame = am

n > 0By induction on n > 0. The case n = 1 follows from Definition 14.2. Supposethat the identity has been established for the numbers 1, 2, , n 1. We willshow that it is still true for n. Indeed,

aman = am(an1a)= (aman1)a= am+n1a= am+n (by Definition 14.2)

By induction, it follows that the identity is true for all n > 0.

n < 0, n = mSince n = m then n + m = 0 and in this case we have am+n = a0 = e. By

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(i), we have aman = amam = e. It follows that aman = am+n.

n < 0, n > mThen m = (m + n) + (n) where both m + n and (n) are positive.

aman = a(m+n)+(n)an

= (am+nan)an

= am+n(anan)= am+ne = am+n

n < 0, n < mIn this case, n = m + [(m + n)] where both m and (n + m) are positive.

aman = am(a1)n

= am(a1)m(m+n)

= am[(a1)m(a1)(m+n)]= am[am(a1)(m+n)]= (amam)(a1)(m+n)

= (a1)(m+n) = am+n

Similar arugument holds for a fixed m < 0 and all integers n. This completesa proof of (ii).

(iii) Similar to (ii) and is left as an exercise.

Remark 14.3

In the case of an group G written with the binary operation +, for n Z+

and a G, one writes na instead of an, where na = a + ... + a (n times),and (n)a = (na) = n(a). The laws corresponding to (ii) and (iii) ofTheorem 14.4 become

ma + na = (m + n)a

andn(ma) = (mn)a.

where m, n Z.

The set of all integral exponents of an element a in a group G forms asubgroup of G as shown in the next theorem.

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Theorem 14.5

Let G be a group and a G. Then the set

< a >= {an : n Z}

is a subgroup of G. In fact, < a > is an Abelian Group.

Proof.

The set < a > is nonempty since a0 = e < a > . Now, let x, y < a > .Then x = an and y = am for some integers n and m. Thus,

xy1 = an(am)1

= anam (by Theorem 14.1(iv))

= anm (by Theorem 14.4(ii))

Since n m Z then xy1 < a > . Thus, by Theorem 7.5, < a > is asubgroup ofG, as required. Thus, < a > is itself a group. Since anam = aman

then < a > is Abelian.

Definition 14.3

The subgroup < a > is called the subgroup generated by a. Any subgroupH of G that can be written as K =< a > is called a cyclic subgroup. Theelement a is called a generator. In particular, G is a cyclic group if thereis an element a G such that G =< a > .

Example 14.1

1. The group Z of integers under addition is a cyclic group, generated by 1(or -1). Thus, (Z, +) is a cyclic group of infinite order.2. Let n be a positive integer. The set Zn of congruence classes of integersmodulo n is a cyclic group of order n with respect to the operation of additionwith generator [1].

The following theorem shows that when powers ofa are equal then the cyclicgroup < a > is of finite order.

Theorem 14.6Let G be a group and a G be such that ar = as for some integers r and swith r = s.

(i) There is a smallest positive integer n such that an = e.

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(ii) at = e if and only ifn|t.

(iii) The elements e,a,a2

, , an1

are distinct and< a >= {e,a,a2, , an1}.

Proof.

(i) Let S = {p Z+ : ap = e}. Assume that r > s. (Ifs > r then interchangethe letters r and s in the following sentences) Since ar = as then aras =asas = e. Thus, r s S so that S = subset ofN. By Theorem 10.1,there is a smallest positive integer n such that an = e.(ii) Suppose first that at = e for some integer t. By the Division Algorithmthere exist integers q and r such that t = nq + r where 0 r < n. Thus,e = at = anq+r = (an)qar = ear = ar. By the definition of n we must haver = 0. That is, t = nq and consequently n|t. Conversely, suppose that n|t.Then t = nq for some integer q. Thus, at = (an)q = e.(iii) First we prove that e,a,a2, , an1 are all distinct. To see this, supposethat au = av with 0 u < n and 0 v < n. Without loss of generality wemay assume that u v. Since au = av then auv = e. By (ii), n|(u v). But0 u v u < n. Thus, we must have u v = 0 or u = v.We prove < a >= {e,a,a2, , an1} by double-inclusions. It is triviallytrue that {e,a,a2, , an1} < a > . Now, let at < a >. By the divisionalgorithm, there exist integers q and r such that t = nq + r with 0 r < n.Thus, at = (an)qar = ar with 0 r < n. That is, at {e,a,a2, , an1}.

This completes a proof of the theorem

Definition 14.4

Let a be an element of a group G. The order of a is the smallest positiveinteger n, if it exists, for which an = e. If such an integer does not exist thenwe say that a has an infinite order. We denote the order of a by o(a).

Example 14.2

1. In (Z4, ), o([2]) = 2.2. In (Q, ) the number 2 has infinite order since 2n = 1 for all positiveintegers n.

We end this section with a theorem that gives the relationship between theorder of a group and the order of an element.

Theorem 14.7

If G is a group and a G then o(a) = | < a > |.

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Proof.

If o(a) = n then by Theorem 14.6 (iii) we have < a >= {e,a,a2

, , an1

}.Thus, | < a > | = n = o(a). If o(a) is infinite then by Theorem 14.6(i) theintegral powers of a are all distinct. Thus, < a > is infinite and o(a) = | |

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