# The Symmetric Groups

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1 Chapter 7 The Symmetric Groups 7. Introduction In the investigation of finite groups the symmetric groups play an important role. Often we are able to achieve a better understanding of a group if we can thin of it as a group of permutations on some set, often one with some geometrical structure. In this chapter we apply much of the theory we have developed so far to study the structure of Sn, the group of permutations on the set {, 2,3,..., n}. We shall show that Sn can be partitioned into two sets, one of which is a subgroup of considerable importance. A partitioning of each of these two sets into conjugacy classes provides us with an important tool which we tae advantage of in our proof that the converse of Lagrange s theorem is not valid. 7.2 The Cayley Representation Theorem We begin with a well-nown theorem which says that any abstract group is isomorphic to a subgroup of some symmetric group. Of course, nowing that a group is contained in some other group does not, in general, tell us much about the group itself, especially if the containing group is big compared with the contained group, as is the case in this representation. Nevertheless, what is important is the idea that an abstract group can be realized as a more concrete object, an idea that has led to a branch of group theory called Representation Theory Theorem (Cayley Representation Theorem) Every group G is isomorphic to a subgroup of some symmetric group. 65

2 66 Abstract Algebra Proof. Let S G be the group of permutations on the underlying set G of the group ( G,.). Define a map f : G SG as follows: f ( g ) = Πg where Π g is the permutation of the set G which g induces by left multiplication; that is, for all x G, Π g ( x) = gx. We must first show that Π is a bijective map on G which will g g establish that Π g is in S G. Π is :. Suppose Π ( x) = Π ( y). Then gx = gy and by cancelling g we get x = y, proving that Π g is :. g Π g is onto G. Let x G Then Π g ( g x) = gg x = x. Next we show that f is a homomorphism. To do this we must prove that f ( gh) = f ( g) f ( h) for all g, h G and this means showing that g Π gh = ΠgΠ h. Naturally, the operation in S G is composition of maps. Now Π ( x) = ( gh) x = g( hx) = Π ( hx) = Π Π ( x) gh g g h for all x G and so Π gh = ΠgΠ h. Therefore, f is a homomorphism. Lastly, we prove f is :. If g er f, then Π g is the identity map on G and so, in particular, Π g ( e) = e. Therefore ge = e, whence g = e and so er f =< e >, thereby establishing that f is :. It now follows that G f ( G) where f ( G ) is a subgroup of S G. Observe that the size of the isomorphic copy of the group G is, in general, very small compared to the size of S G. For example, if the order of G is 6, then the order of S G is 6! = 720. The subgroup (of order 6) of S isomorphic to G is inside a group 20 times its size! G

3 The Symmetric Groups Example We realize the Klein 4-group K 4 as a permutation group. Recall that K4 = { e, a, b, c} where the square of any element is e and the product of any two distinct nonidentity elements is the third element. We have:

4 68 Abstract Algebra e a b c e a b c e a b c e a b c Π e = I = ; Π a = ; Π b = ; Π c =. e a b c a e c b b c e a c b a e If we set e =, a = 2, b = 3, c = 4, the permutations are [234], [243], [342], and [432] form a subgroup of S Permutations as Products of Disjoint Cycles In order to study S n in detail, we need to introduce notation for the elements of S n which conveys at a glance the nature of the elements. The notation we have been using so far doesn t do this. For example, [234] is a rather tame element of S4 since it moves none of,2,3,4. On the other hand, [234] is quite dynamic since every number is moved by it. The eye, however, does not immediately distinguish between the two Definition and notation A permutation π such that: π ( ai ) = ai +, i, π ( a ) = a is denoted by ( a a2 a3... a ) and is called an n- cycle or a cycle of length n. Two cycles without any common symbols are said to be disjoint. The identity permutation is written as () although, strictly speaing we should write it as ()(2)(3)...( n ), but this is clearly too cumbersome. Note. The n- cycles ( a a2 a3... a ), ( a2 a3 a4... a a ),...,( a a a2... a ) all represent the same permutation Example (2)(345)(6)(7) is the permutation which maps to 2, 2 to, 3 to 4, 4 to 5, 5 to 3 and leaves 6 and 7 fixed. We would say that this permutation is a product of disjoint cycles. In general the -cycles are omitted. We would therefore write (2)(345)

5 The Symmetric Groups 69 when it is clear that the permutation is an element of S7. In short, each omitted symbol is assumed to be mapped onto itself but we must indicate which S n the permutation belongs to Theorem Every permutation of Sn is a product of disjoint cycles. Proof. Let g S and for some i {,2,3,..., n} consider the sequence n 0 2 i = g ( i), g ( i) = g( i), g ( i),..., g ( i),... There must be a repetition in this sequence since each g m (i) {,2,3,,n} for all m. Let be the first positive integer such that g ( i ) is a repetition of some preceding term of the sequence, so that 0 2 i = g ( i), g( i), g ( i),..., g ( i) p are distinct while g ( i) = g ( i) for some nonnegative p. Applying p g p 0 to each side of this equation, we get g ( i) = g ( i) = i. Now p 0 p and g ( i) is a repetition of a preceding term of the sequence, namely, i. Therefore, by the minimality of, p cannot be strictly less than. Therefore, p = 0, showing that the first repetition is i, i.e., g ( i) = i. We can illustrate the foregoing by a diagram: i m g ( i) g( i ) 3 g ( i ) g 2 ( i ) Fig. 7..

6 70 Abstract Algebra We claim that 2 { g ( i) m Z } = { i, g( i), g ( i),..., g ( i)}. It is obvious that the right-hand side is contained in the left-hand side. Conversely, for any integer m, m = q + r where 0 r < by the Division Algorithm and so m q+ r r q r g ( i) = g ( i) = g [( g ) ( i)] = g ( i) q since ( g ) ( i) = i (why? See the diagram above), and so our assertion is proved. (The subset 2 { i, g( i), g ( i),..., g ( i)} orbit of i under g and is denoted by Og ( i ).) of {, 2,3,..., n } is called the Next, pic any j in {,2,..., n } but not in O ( i ) (if such a j exists). Proceed as above to get 2 3 t { j, g( j), g ( j), g ( j),..., g ( j)} t where g ( j) = j. We claim that O ( i) O ( j) = φ.. Indeed, suppose u v g ( i) = g ( j). g u v Then g ( i) = j O ( i), a contradiction since j O ( i). g g Continue producing orbits as above until all the elements of {, 2,..., n } have been exhausted. We obtain a collection of non-overlapping subsets of {,2,..., n } whose union is {,2,..., n}. It is now clear that 2 2 t 2 v ( )( ) ( ) g = i g( i) g ( i).. g ( i) 牋 j g( j) g ( j).. g ( j)...? l g l) g ( l).. g ( l). g g Let us tae a specific example of our discussion above Example Consider g = [ ] S9. We compute the orbits of g. The orbit of is {, g () = 2} since g 2 () = g(2) =. Similarly, the orbits of 3 and 6 are, respectively, {3,4,5} and {6,7,8,9}. The following diagram exhibits quite clearly the behavior of g

7 The Symmetric Groups 7 Figure 7.2 We can indicate the behavior of g in the first orbit by writing (2). We tae this to mean that g maps to 2 and 2 bac to. As for the second orbit, we write (345) which conveys that g maps 3 to 4, 4 to 5 and 5 bac to 3. For the last orbit, g maps 6 to 7, 7 to 8, 8 to 9 and 9 bac to 6. Therefore we can write g as (2)(345)(6789). When computing with permutations expressed as products of disjoint cycles, the next theorem is useful Theorem Disjoint cycles commute. Proof. Suppose α, β are disjoint cycles. Then α fixes all symbols moved by β and vice-versa. Symbols which do not appear in either are fixed by both αβ and βα. Suppose now that j is a symbol appearing in α (and therefore not in β ). Then αβ ( j) = α( j) and, since α ( j ) does not appear as a symbol in β, it follows that β[ α( j)] = α( j). Hence αβ ( j) = α( j) = βα( j). A similar argument shows that if is a symbol appearing in β (and so not in α ), we have αβ ( ) = β ( ) = βα( ). Therefore, αβ = βα Corollary If α, β,..., π are pairwise disjoint cycles, then, for all integers m, ( αβ... π ) m = α m β m... π m. Proof. Exercise. (Use the previous theorem and induction). Before proving the next theorem, we give an example to motivate it.

8 72 Abstract Algebra Example What is the order of the permutation (2)(345)(6789)? Squaring, we get [(2)(345)(6789)] = (2) (345) (6789) = (354)(68)(79). 2 Observe that (2) = () and so (2) disappears. Cubing yields (2) (345) (6789) = (2)(6987). This time (345) disappears since (345) = (). This suggests that to suppress an m -cycle we must raise the cycle to a multiple of m. Therefore, to suppress all symbols occurring in the given permutation, we must raise it to the power 2 which is the least common multiple of the lengths of the cycles and this is the smallest such power. Therefore the order of (2)(345)(6789) is Lemma The order of an m-cycle is m. Proof. Suppose the m-cycle is α = ( j0 j j2... jm ). Then ( j ) = j where for this proof only, s + mod m means the α s s+ mod m unique integer t such that t s + mod m, 0 t < m. For α to be the identity map, it is necessary and sufficient for to be the smallest positive integer for which s + smod m for all s, 0 s < m; that is, subtracting s from each side of the congruence, we want the smallest positive integer such that 0mod m. Obviously, = m fits the bill and so the order of an m -cycle is m.

9 The Symmetric Groups Theorem Suppose the permutation π is the product of the disjoint cycles α, =,2,..., s where α i is an m i -cycle. Then the order of π is i i lcm{ mi i =,2,..., s}. Proof. Let n = lcm { mi { i =,2,..., s}. By Corollary and the n n n n preceding lemma, π = α α2... α s = () and so π divides n.. On the π π other hand, π = () and so α i = () for all i, i =,2,..., s which implies that m divides π for all i, i =, 2,..., s by Lemma Therefore i n divides π and so π = n Example Express the permutation α = (34)(9263)(4735)(789) as a product of disjoint cycles and find its order. Note that the cycles in α are not disjoint. To express α as a product of disjoint cycles we apply α to each of the symbols appearing in its cycles. We get α(7) = 8, α(8) = 2, α(2) = 6, α(6) = 4, α(4) = 7 and the cycle is complete. Hence, one of the disjoint cycles is (78264). The symbol does not appear in this cycle and so we apply α to and proceed as above to obtain the cycle (935). Therefore α = (78264)(935) and so its order is 20, the lcm of 5 and 4, by Theorem Remar. The decomposition of a permutation as a product of disjoint cycles enables us to write down its inverse immediately merely by writing the cycles bacwards. For example, in the previous example, α = (46287)(539). The reader should provide a proof of the general statement.

10 74 Abstract Algebra 7.4 Odd and Even Permutations As mentioned in the Introduction, we show that the elements in Sn fall into two classes, the odd permutations and the even permutations. There are a number of proofs of this result which is quite tricy. We have chosen the one presented because it puts homomorphisms to use Definition A permutation which interchanges two symbols and leaves all others fixed is called a transposition Lemma Every m-cycle is a product of m transpositions. Proof. It is a simple matter to chec that ( j j2 j3... jm) = ( j jm )( j jm )( j jm 2)...( j j3)( j j2), a product of m transpositions. (Remember that the right-most map is applied first.) Theorem. Every permutation is a product of transpositions. Proof. Each permutation is a product of disjoint cycles and each cycle, by the lemma, is a product of transpositions. Therefore, each permutation is a product of transpositions. Remar. If a permutation is expressible as a product of disjoint cycles of lengths r, s, t,..., w then it can be expressed as a product of

11 The Symmetric Groups 75 ( r ) + ( s ) + ( t ) ( w ) transpositions since a cycle of length is a product of transpositions, by Lemma Example (23)(45)(6789) = (3)(2)(45)(69)(68)(67). Remars. (i) Observe that, in general, the transpositions are not disjoint. (ii The decomposition of a permutation as a product of transpositions is not unique. For example, (23)(45)(6789) = (3)(2)(45)(69)(68)(67) = (2)(23)(45)(47)(46)(49)(48)(47). Note that even the number of transpositions in the two decompositions is not the same. However, the number of transpositions in each decomposition is even. (iii) As another example we have (23)(4567) = (3)(2)(47)(46)(45) = (23)(34)(4)(34)(47)(46)(45). Again, although the number of transpositions in each of the two decompositions is different, each is odd. (iv) We shall eventually prove that if a permutation decomposes as a product of an even (respectively, odd) number of transpositions, then any decomposition of that permutation will have an even (respectively, odd) number of transpositions. We shall then be able to classify a given permutation as odd or even. To prove the assertion in (iv), we need to introduce some notation and prove two lemmas. Notation. Let π be a permutation and let π ( i) π ( j) S( π ) = i j P where P is the set of all (unordered) pairs { i, j} {,2,3,..., n}. We see that S is a map from S n to the rational numbers.

12 76 Abstract Algebra Remar. Observe that the order of the pairs i, j is immaterial since π ( i) π ( j) π ( j) π ( i) =. i j j i Example Let π = (2)(34). Then there are 6 subsets of {,2,3,4} of size 2, namely, {,2}, {,3}, {,4}, {2,3}, {2,4} and {3,4}. Then [ π () π (2)]...[ π (2) π (3)][ π (2) π (4)][ π (3) π (4)] S( π ) = [ 2] [ 3] [2 3] [2 4] [3 4] [2 ][2 4][2 3][ 4][ 3][4 3] = = [ 2][ 3][ 4][2 3][2 4][3 4] In practice, when computing S( π ), it is convenient to write π as follows: n. π () π (2) π (3) π (4)... π ( n) Then the pairs can be listed as {,2}, {,3}, {,4},...,{, n}; {2,3}, {2,4},...,{2, n};...;{ n, n} and S( π ) easily computed. In the listing of the pairs, we first list all pairs with as the smaller number of the pair, then all pairs with 2 as the smaller number, all pairs with 3 and so on Lemma If α and β are permutations, then S( αβ ) = S( α) S( β ). Therefore, S is a homomorphism from nonzero rational numbers. S n to the multiplicative group of

13 The Symmetric Groups 77 Proof. αβ ( i) αβ ( j) S( αβ ) = and i j observe that α can be written as P S( β ) = P β ( i) β ( j). We i j β () β (2) β (3) β (4)... β ( n) αβ () αβ (2) αβ (3) αβ (4)... αβ ( n) and so, we can write αβ ( i) αβ ( j) S( α) =. β ( i) β ( j) Multiplying S( α ) and S( β ) together, we get S( α ) S( β ) = P P αβ ( i) αβ ( j) β ( i) β ( j) P = S( αβ ) β ( i) β ( j) = i j since the β ( i) β ( j) cancel. Therefore, S( αβ ) = S( α) S( β ). P αβ ( i) αβ ( j) i j Corollary If α, α2, α3,..., α are permutations, then S( α α α... α ) = S( α ) S( α ) S( α )... S( α ). 2 3 n 2 3 Proof. Use the above result and induction. The final result we need is that the image of a transposition under S is. n Lemma If τ is a transposition, then S( τ ) =. Proof. Without loss of generality we may assume τ = (2). Then

14 78 Abstract Algebra n S( τ ) = S = n (2 )(2 3)...(2 n)( 3)...( n)(3 4)(3 5)...([ n ] n) =. ( 2)( 3)...( n)(2 3)...(2 n)(3 4)(3 5)...([ n ] n) Theorem If, α = τ τ 2 τ 3..τ p = σ σ 2 σ 3..σ q where the τ 's and σs are p q transpositions, then ( ) = ( ), or, in other words, either both p and q are even or both odd. Proof. Applying S to α, we have S( α) = S( τ τ... τ τ ) = S( σ σ... σ σ ). 2 p p 2 q q By Corollary 7.4.7, S( τ ) S( τ ) S( τ )... S( τ ) = S( σ ) S( σ ) S( σ )... S( σ ). 2 3 p 2 3 By Lemma 7.4.8, S( τ ) = S( σ ) = for all i, i p and for all j, j q. i p j q Therefore, ( ) = ( ) from which we deduce that either both p and q are odd or both even. Remar. Note that we have incidentally proved that the map S is actually from the group to the multiplicative subgroup {, } of nonzero reals. q Definition A permutation is said to be even if it is expressible as a product of an even number of transpositions. Otherwise, it is said to be odd. The property of being odd or even is the parity of the permutation.

15 The Symmetric Groups Corollary The set of all even permutations of Sn, n >, is a normal subgroup of S n of order n! called the alternating group of degree n and denoted by A 2 n. Proof. By the First Isomorphism Theorem, Lemma and Theorem S n above, {, } er S. But er S = A n and so Sn {, } A n and A n is normal of index 2 in S n. Hence there are only two cosets of A n in S n, one consisting of all the even permutations and the other all the odd permutations. It now follows that A n! n = S = n Example We list the elements of S 4 according to their cycle structure. We say that two permutations have the same cycle structure if in each decomposition as a product of disjoint cycles, the number of cycles is the same and the number of cycles of a given length is the same in each. For example, (23)(45)(6789) and (689)(2345)(7) have the same cycle structure. In the table below the number of elements of a given cycle structure is calculated using fundamental principles of counting. For example, to compute the number of elements having the same cycle structure as (23) (considered as an element of S 4 ), we argue as follows: there are 24 = different ordered triples since there are 4 ways of choosing the first position, 3 ways of choosing the second and 2 ways of choosing the third. These 24 ordered triples, however, do not yield 24 different permutations. Indeed, the ordered triples 23, 23 and 32 give

16 80 Abstract Algebra rise to the same permutation. If we collect together those ordered triples which yield the same permutation, each such collection will contain three ordered triples and so, there will be 24 = 8 collections. Therefore, there 3 are 8 permutations with the same cycle structure as (23) in S 4 Table 7.. Cycle structure of S 4 Cycle Number Order Parity Structure () Even (2) 6 2 Odd (23) 8 3 Even (234) 6 4 Odd (2)(34) 3 2 Even 7.5 Conjugacy Classes of a Group We tae a brief excursion away from symmetric groups to develop some more theory which we need to further our study of the symmetric groups. We first define a relation called conjugacy and show that it is an equivalence relation whose classes we call the conjugacy classes of the group. This decomposition of a group maes its structure more transparent and we shall see later on that we shall mae full use of it Definition An element x of a group G is said to be conjugate in G to an element y if there exists an element g of the group such that y = gxg. We write x ; y and read: x is conjugate to y. To conjugate x by g means g to compute gxg x = x (some authors mean x gx = g for conjugation by x).

17 The Symmetric Groups Theorem Conjugacy is an equivalence relation on G whose equivalence classes are called the conjugacy classes of G. The conjugacy class of a is denoted by K( a ). Proof. (i) (Reflexivity) For all x G, x ; x since x = exe. (ii) (Symmetry) Suppose x ; y. Then there exists g G such that y = gxg. Multiplying this equation on the left by g and on the right by g, we get x = g y( g ) and so y ; x. (iii) (Transitivity) Suppose that x ; y and y ; z. Then there exist g and h in G such that y = gxg and z = hyh. Hence z = hgxg h = ( hg) x( hg) and so x ; z. Therefore ; is an equivalence relation. Remar. The conjugacy classes of an abelian group are all singleton sets so conjugacy turns out to be quite uninteresting in the abelian case. When are two elements conjugate in Sn? The answer turns out to be quite esthetically pleasing. We begin with a lemma Lemma Let π be a permutation. Then (*) π ( a a... a ) π = ( π ( a ) π ( a )... π ( a )). 2 2 Proof. Applying the left-hand side of equation above to π ( a ), we get π ( aa2... a ) π [ a π ( ) ] = π ( aa2... a )[ a ] = π ( a2). In general, applying the left-hand side to π ( a j ), j <, we get π ( a j + ) and when j =, 2 = 2 = π ( a a... a ) π [ π ( a )] π ( a a... a )[ a ] π ( a ). Therefore, one of the nontrivial cycles of the permutation π ( aa2... a ) π is ( π ( a ) π ( a2)... π ( a )).

18 82 Abstract Algebra The permutation on the right-hand side of (*) moves the symbols π ( a ), j and no others. We need only show that the same is true j of π ( aa2... a ) π and the result will be proved. Suppose b { π ( a ), π ( a2),..., π ( a )}. Then π ( b) { a, a2,..., a }(why?). Hence ( aa2... a )[ π ( b)] = π ( b) and so 2 2 π ( a a... a ) π [ b] = π{( a a... a )[ π ( b)]} = ππ ( b) = b Theorem Two permutations are conjugate in S n if, and only if, they have the same cycle structure. Proof. Suppose first that the two permutations λ and µ have the same cycle structure, say λ = τ τ... τ, µ = σ σ... σ 2 2 where τ j = ( a j a j2 a j3... a jt ) and σ j j = ( bj bj2 bj3... bjt ). The number of j symbols appearing in each of λ and µ is symbols fixed by each is j= j j= t j and so the number of n t. Let ρ be any bijection from the symbols fixed by λ to the symbols fixed by µ and define a permutation π as follows: bjs if x = a js; π ( x) = ρ( x) otherwise. In the decomposition of λ as a product of disjoint cycles, insert π π between each pair of consecutive τ s and multiply λ by π on the left and by π on the right. We get 2 3 πλπ = ( πτ π )( πτ π )( πτ π )...( πτ π ).

19 By Lemma 7.5.3, proving that πλπ = µ. The Symmetric Groups 83 =, πτ jπ σ j for all j, j Conversely, suppose λ and µ are conjugate, say µ = γλγ and let the decomposition of λ as a product of disjoint cycles be as in the first part of the proof. Then µ = γλγ = ( γτ γ )( γτ γ )( γτ γ )...( γτ γ ). By the lemma, each Moreover, the cycles j 2 3 γτ γ is a cycle of the same length as that of τ j. γτ jγ, j =,2,..., are disjoint (why?) and so λ and µ have the same cycle structure Example Find a permutation π such that µ = πλπ where λ = (,2)(3,5,0)(2,4,6,7) and µ = (0,)(2,4,3)(9,8,,5). The symbols fixed by λ are8,9 and ; those fixed by µ are 6,7 and 2. We define any bijection between these two sets of symbols, say 8 6, 9 7, 2. Next, place the decomposition of µ under the decomposition of λ so that cycles of the same length line up under each other. We get: This array defines a map which sends a symbol in the top line to the symbol immediately below it. This map, together with the bijection defined above between the symbols fixed by the two permutations, yields the permutation (,0,3,,2,9,7,5,4,8,6), which is the required permutation π. Note that since we have a choice for the bijection between the elements fixed by each permutation, the solution above is one of six possible answers. There are in fact many more which we obtain by

20 84 Abstract Algebra cyclically permuting the symbols in each cycle and following the recipe above. Lagrange s Theorem (Corollary 6.2.7) states that the order of a subgroup of a finite group divides the order of the group. However, given a divisor of the order of a group, there does not necessarily exist a subgroup of order the given divisor. We shall give an example of a group of order 2 which contains no subgroup of order 6. To do this, we first prove a couple of simple results Theorem If H is a subgroup of G of index 2 in G, then H is normal in G. Proof. We need to show that, for all g G, gh = Hg. If g is an element of H, gh = Hg = H. Suppose g is not an element of H. Then, since H is of index 2, we have G = H gh= H Hg. Since this is a disjoint union, the complement of H in G is gh and also Hg. Hence, gh = Hg Theorem If H is a normal subgroup of G and a is an element of H, then the full conjugacy class of a in G is contained in H. Proof. Every element of the conjugacy class of a is expressible as gag for some g in G. But a H implies gag ghg = H, the last equality being a consequence of the normality of H and Theorem Therefore, the full conjugacy class of g is contained in H Theorem The alternating group A4 of Example has no subgroup of order 6.

21 The Symmetric Groups 85 Proof. Let us suppose that there exists such a subgroup H. Since A 4 = 2, this hypothetical subgroup is of index 2 in A 4 and so, by above, H is normal in G. Now A 4 contains the identity, eight 3 -cycles and the elements (2)(34), (3)(24) and (4)(23). Hence our subgroup H must contain at least one 3 -cycle, say, ( abc ). By 7.5.7, H must contain all conjugates of ( abc ) in A 4. Note that if we new that H was normal in S 4, it would have to contain all eight 3 -cycles by and 7.5.4, which would be impossible since the order of H is only six. In this manner, we would have proved quite easily that H does not exist. However, we now only that H is normal in A 4, and so we have to wor a little harder to force more elements in H. We compute some of the conjugates of ( abc ), being careful to conjugate only by elements of A4 only: =. ( abd)( abc)( abd) ( bdc); ( bad)( abc)( bad) = ( dac); ( bdc)( abc)( bdc) = ( adb). Thus H must contain { ( abc), ( bdc), ( dac), ( adb )} as well as the identity (). Therefore, five elements are already accounted for. Also, since H is a subgroup, it must contain the inverse of each of its elements, and so must contain: ( abc) = ( acb), ( bdc) = ( bcd), ( dac) = ( dca), ( adb) = ( abd ). But this means that H must contain at least nine elements, which is absurd since the order of H is only six. Therefore, H does not exist. (See Problem 9 for another proof). 7.6 Exercises ) Follow the steps in the proof of Theorem 7.2. in the particular case that the group whose representation is being computed is a cyclic group of order 5. 2) Compute the order and parity of the following permutations:

22 86 Abstract Algebra (i) (347)(235)( 4735)(29847); (ii) (2)(53)(243)(36); (iii) (357)(2463)(234567); (iv) (23)(342)(546)(36). 3) Express each of the permutations in 2) as products of transpositions in two different ways. 4) Draw up a table for S 5 similar to the one in Example ) Let m and n be relatively prime positive integers. (i) If = rm + sn, prove that gcd( r, n) = gcd( s, n) =. (ii) If b is an element of a group of order mn where m and n are as rm sn above, prove that b = n and b = m. Hence prove that b is expressible as a product of an element of order m and an element of order n. (iii) Express (23456) as a product of an element of order 2 and an element of order 3 and write each of these elements as a product of disjoint cycles. 6) If π is a permutation such that (i) π (234)(56) = (3726)(4), express π as a product of disjoint cycles; (ii) π (234)(56) π = (3726)(4), express π as a product of disjoint cycles. (There is more than one answer.) 7) In S 3 find an element of order 42 and one which moves all symbols and is of order 20. 8) Prove that the smallest subgroup of S n containing (2) and (23.n) is S n itself. [Hint: Show that any subgroup containing the two given elements must contain all transpositions.] 9) If σ and τ are transpositions, show that στ can be expressed as a product of (not necessarily disjoint) 3-cycles.

23 The Symmetric Groups 87 0) Let G be a finite group and g an element of G. Recall (see Problem 9, Chapter 5) that the centralizer of g in G is the subgroup CG ( g) = { x G xg = gx}. To simplify notation we shall denote this subgroup by C. Prove that the number of elements in the conjugacy class K( g ) of g in G is G : C ( g ), the index of C ( g ) in G. G [Hint: Define a map ϕ from the set of left cosets of C in G to K( g ) by ϕ( xc) = xgx. First show that ϕ is well defined. This has to be done since it appears that the image of a coset under ϕ depends on the coset representative. Then show that the mapϕ is bijective.] ) What is the order of: (i) C S ((2)(34))? 5 (ii) CS n ((23... n ))? 2) Prove that the conjugacy class of (2345) in A 5 is one half the size of the conjugacy class of (2345) in S 5 but that the conjugacy class of (23) in A 5 is of the same size as the conjugacy class of (23) in S 5. 3) Prove that if n > 2, then every element of A n is expressible as a product of 3-cycles. 4) (Hard!) Prove that if H is a nontrivial normal subgroup of A 5, then H = A 5. 5) Prove Z(S n ) = < () >. (See Problem 3, Chapter 5). 6) Prove that if τ, τ 2, τ 3 are transpositions, then τ τ 2 τ 3 (). 7) Prove that if τ, τ 2 ττ 2 is 2 or 3. G are distinct transpositions, then the order of

24 88 Abstract Algebra ) If the permutation 3 2?? must the images of 4 and 5 be? in S 9 is even, what 9) Give another proof of the fact that A4 has no subgroup of order 6 as follows: (i) If H is a subgroup of A4 of order 6, then H is a normal subgroup of A 4; 2 (ii) Show that for each g A4, g H; (iii) Prove then that H would have too many elements. 20) If α = (3579)(246)(8,0) and α m is a 5-cycle, what can you say about m? 2) Describe the conjugacy classes of A 6 and compute the number of elements in each class. 22) Show that a permutation of odd order must be of even parity. 2 23) If α S 3 and α = (,0,,6,3,2,2,9,5,3,7,4,8) what is α? 24) Show that if H is a subgroup of S n, then either every element of H is an even permutation or exactly half of them are even. 25) If σ and τ are permutations, prove that στ and τσ have the same cycle structure. 27) Does the symmetric group S 7 contain an element of (i) order 5? order 0? order 5? (ii) What is the largest possible order of an element of S7? 28) A dec of cards numbered to 2n is in the order, 2,,2n. A perfect shuffle is performed so that the cards end up in the order n +,, n + 2, 2, n + 3, 3,..., 2 n, n. How many perfect shuffles must be performed to return the cards to their initial order?

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