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Arkansas Tech University

MATH 4033: Elementary Modern AlgebraDr. Marcel B. Finan

6 Permutation Groups

Let Sbe a nonempty set and M(S) be the collection of all mappings fromS into S. In this section, we will emphasize on the collection of all invertiblemappings fromSintoS. The elements of this set will be called permutationsbecause of Theorem 2.4 and the next definition.

Definition 6.1

Let S be a nonempty set. A one-to-one mapping fromS onto Sis called apermutation.

Consider the collection of all permutations on S. Then this set is a groupwith respect to composition.

Theorem 6.1

The set of all permutations of a nonempty set Sis a group with respect to

composition. This group is called the symmetric group on Sand will bedenoted by Sy m(S).

Proof.

By Theorem 2.4, the set of all permutations on S is just the set I(S) of allinvertible mappings from S to S. According to Theorem 4.3, this set is agroup with respect to composition.

Definition 6.2

A group of permutations , with composition as the operation, is called apermutation group on S.

Example 6.1

1. Sym(S) is a permutation group.2. The collectionL of all invertible linear functions from Rto Ris a permu-tation group with respect to composition.(See Example 4.4.) Note that Lis

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a proper subset of Sym(R) since we can find a function in Sy m(R) which

is not in L, namely, the function f(x) = x3

. This example shows that, ingeneral, a permutation group on Sneeds not contain all the permutationson S.

Example 6.2

Let S={1, 2, 3}. There are six permutations on S. We will represent thesepermutations using the two-row formas follows:

1=

1 2 31 2 3

, 2 =

1 2 32 3 1

, 3=

1 2 33 1 2

4=

1 2 32 1 3

, 5 =

1 2 33 2 1

, 6=

1 2 31 3 2

In composing permutations we always follow the same convention we use incomposing any other mappings: read from right to left. Thus,

1 2 33 1 2

1 2 33 2 1

=

1 2 32 1 3

That is, 1 3 2, 2 2 1, 3 1 3.

Example 6.3

Let S = {1, 2, 3}. Then Sy m(S) = {1, 2, 3, 4, 5, 6}, where the s aredefined in Example 6.2. Lets construct the Cayley table for this group.

1 2 3 4 5 61 1 2 3 4 5 62 2 3 1 5 6 43 3 1 2 6 4 54 4 6 5 1 3 25 5 4 6 2 1 36 6 5 4 3 2 1

Notice that the permutation

1 = 1 2 31 2 3

is the identity mapping ofS ym(S). Moreover,

1 2 32 3 1

1

=

1 2 33 1 2

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Thus, the inverse of an element is obtained by reading from the bottom entry

to the top entry rather than from top to bottom:if 1 appears beneath 3 in 2then 3 appears beneath 1 in 12 .We will denote the above group by S3. In general, ifS={1, 2, , n} thenthe symmetric group on Swill be denoted by Sn.

The number of elements ofSn is found in the following theorem.

Theorem 6.2

The order ofSn is n!,where 0! = 1! = 1 and n! =n(n 1)(n 2) 2 1.

Proof.

The proof involves the following counting principle: If a decision consists of

two steps, if the first step can be done inr different ways and the second stepcan be done ins different ways then the decision can be made inrsdifferentways.The problem of computing the number of elements ofSn is the same as theproblem of computing the number of different ways the integers 1, 2, , ncan be placed in the nblanks indicated(with each number used only once)

1 2 3 n

Filling the blanks from the left, we see that the first blank can be filled withn different ways. Once this is completed, the second blank can be filled inn 1 ways, the third in n 2 ways and so on. Thus, by the principle ofcounting, there are n(n 1)(n 2) 2 1 =n! ways of filling the blanks. Inconclusion,|Sn|= n!

Now, since S1 = {(1)} then S1 with respect to composition is commuta-tive. Similarly, since (1)(12) = (12)(1) then S2= {(1), (12)} is also Abelian.Unfortunately, this is not true anymore for |S|> 2.

Theorem 6.3

Sn is non-Abelian for n 3.

Proof.All that we need to do here is to find two permutations and in Sn withn 3 such that = . Indeed, consider the permutations

=

1 2 3 4 5 n1 3 2 4 5 n

and =

1 2 3 4 5 n3 2 1 4 5 n

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Then

=

1 2 3 4 5 n2 3 1 4 5 n

and =

1 2 3 4 5 n3 1 2 4 5 n

so that =

Cycle Notation for Permutations

The cycle notation for permutations can be thought as a condensed way towrite permutations. Here is how it works.Let Sn be the permutation

(a1) =a2, (a2) =a3, , (ak) =a1

and (ai) = ai for i = k+ 1, , n, where a1, a2, , an {1, 2, 3, , n}.That is, follows the circle pattern shown in Figure 6.1

Figure 6.1

Such a permutation is called a cycle of lengthk or simply a k-cycle. We willwrite

= (a1a2a3 ak) (1)

Let us elaborate a little further on the notation employed in (1). The cyclenotation is read from left to right, it says takesa1 into a2, a2 into a3, etc.,and finally ak, the last symbol, into a1, the first symbol. Moreover, leavesall the other elements not appearing in the representation (1) fixed.

Example 6.4

The permutation

=

1 2 3 4 5 6 71 6 3 7 5 4 2

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can be represented as a 4-cycle

= (2647).

Note that one can write the same cycle in many ways using this type ofnotation.

= (2647)= (6472)= (4726)= (7264)

Remark 6.1

A k-cycle can be written in k different ways, since

(a1a2. . . ak) = (a2a3. . . aka1) = = (aka1. . . ak1).

Example 6.5

It is easy to write the inverse of a cycle. Since (ak) = ak+1 implies1(ak+1) = ak, we only need to reverse the order of the cyclic pattern.For example,

(2647)1 = (7462).

Example 6.6

Multiplication of cycles is performed by applying the right permutation first.Consider the product in S5

(12)(245)(13)(125)

Reading from right to left

12 2 4 4

so 14.Now

44 4 5 5

so 45.Next

51 3 3 3

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so 53.

Then 33 1 1 2

so 32.Finally

25 5 2 1

so 21. Since all the elements ofA= {1, 2, 3, 4, 5} have been accountedfor, we have

(12)(245)(13)(125) = (14532).

Remark 6.2

A 1-cycle of Sn is the identity of Sn and is denoted by (1). Of course,(1) = (2) = (3) = = (n).

Now not all permutations are cycles; for example, the permutation

=

1 2 3 4 5 6 7 82 3 1 5 6 7 4 8

is not a cycle. However, one can check easily that

= (123)(4567)

This suggests how we may extend the idea of cycles to cover all permutations.

Definition 6.3

If and are two cycles, they are called disjoint if the elements movedby one are left fixed by the other, i.e., their cycle representations containdifferent elements of the set S={1, 2, 3 , n}.

Example 6.7

The cycles (124) and (356) are disjoint whereas the cycles (124) and (346)are not since they have the number 4 in common.

Theorem 6.4

Ifand are disjoint cycles then = .

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Proof.

Indeed, since the cycles and are disjoint, each element moved by isfixed by and vice versa. Let = (a1a2 as) and = (b1b2 bt) where{a1, a2, , as} {b1, b2, , bt}= .

(i) Let 1 k s. Then

()(ak) =((ak)) =(ak) =ak+1

and()(ak) =((ak)) =(ak+1) =ak+1.

(ii) Let 1 k t. Then

()(bk) =((bk)) =(bk+1) =bk+1

and()(bk) =((bk)) = (bk) =bk+1.

(iii) Let 1 m n and m {a1, a2, , as, b1, b2, , bt}. Then

()(m) =((m)) = (m) =m

and()(m) =((m)) = (m) =m.

It follows from (i), (ii), and (iii) that = .

Theorem 6.5

Every permutation ofSn is either a cycle or can be written uniquely, exceptfor order of cycles or the different ways a cycle is written, as a product ofdisjoint cycles.

Proof.

The proof is by induction on n. Ifn = 1 then there is only one permutation,

and it is the cycle (1).Assume that the result is valid for all sets with fewer than n elements. Wewill prove that the result is valid for a set with n elements.Let Sn. If = (1) then we are done. Otherwise there exists a positiveinteger m such that m1(1) = 1 and m(1) = 1. For example, if =

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(145) S5 then m = 3 since (1) = 4, 2(1) = ((1)) = (4) = 5, and

3

(1) =(2

(1)) =(5) = 1.LetQ= {1, (1), 2(1), . . . , m1(1)}.

If Q = {1, 2, . . . , n} then is the cycle = (1(1)2(1) . . . m1(1)). IfQ=S, where S={1, 2, , n}, then

= (1(1)2(1) . . . m1(1))

whereis a permutation on the set S Q= {t S: t Q}. Since this sethas order smaller than n, then by the induction hypothesis can be writtenas a product of disjoint cycles, say, =12. . . k. Thus,

= (1(1)2(1) . . . m1(1))12. . . k

But this says that is expressed as a product of disjoint cycles. This com-pletes the induction step, and establishes the result for all n

Example 6.8

Let S={1, 2, ..., 8} and let

=

1 2 3 4 5 6 7 82 4 6 5 1 7 3 8

Start the first cycle with 1 and continue until we get back to 1, and thenclose the first cycle. Then start the second cycle with the smallest numbernot in the first cycle, continue until we get back to that number, and thenclose the second cycle, and so on to obtain

= (1245)(367)(8)

It is customary to omit such cycles as (8), i.e., elements left fixed by , andwrite simply as

= (1245)(367)

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