Math 375
Week 5
5.1 Symmetric Groups
The Group S
X
Now let X be any non-empty set. Let
S = f j : X ! X and f is b oth surjective and injective g:
X
Notice that if ; 2 S , then : X ! X and is surjective and injec-
X
tiveby previous arguments. That is, the comp osition 2 S .Thus
X
comp osition is a binary op eration on S .
X
THEOREM 1 S is a group under comp osition. S is called the Symmteric Group
X X
on X .
PROOF i We just checked closure. ii Asso ciativity: mentioned last time
see text Chapter 0 . iii The identityinS is i b ecause X x =
X X
i x = x , that is, i = . Similarly i = . iv Evidently
X X X
1
the inverse of is whichwe know exists b ecause is injective and
1 1
surjective and x = x = i x and x = x = i x.
X X
This last example shows just how general the group concept is. Sym-
metric groups are extraordinarily imp ortant. Arthur Cayley as in Cay-
ley table showed that every group is the subgroup of some symmetric
group. So if you understand symmetric groups completely, then you un-
derstand all groups! We can examine S for any set X .For example
X
if X = R, then examples of of elements in S are i , f : R ! R by
R R
a ! a +1, g : R ! R by a ! a=2, and so on. It is clear that S is
R
in nite.
The individual elements of S are often called p ermutations of
X
the elements of X . This will b e the next big topic wecover. 1
2 Math 375
The Symmetric Group on n Elements: S
n
To make matters simpler, we will study symmetric groups of nite sets.
For example, if X is a set of n elements, then wemayaswell lab el
the elements of X as f1; 2;:::;ng.We usually denote the symmetric
group on n elements by S .
n
Nowany elementorp ermutation in S is an injective and sur-
n
jective function from the set of the rst n integers to itself. It merely
shues these elements around. Consequently it can b e represented as
atworow matrix in which the rst row represents the input and the
second represents the corresp onding output.
EXAMPLE 1 List all the elements of S .
3
1 2 3 1 2 3 1 2 3
2
= ; = ; =
1 2 3 2 3 1 3 1 2
1 2 3 1 2 3 1 2 3
2
= ; = ; =
1 3 2 3 2 1 2 1 3
It should b e clear what the identity element is. Howdowe get the
inverse of any element? Merely exchange rows, and re-order. How
do we take the pro duct or comp osition of two such elements in S ?
n
In comp osing functions always remember to work from the right to the
left whichisbackwards from what we read. This is a hold over from
comp osition of functions, which is what we implicitly are doing here.
1 2 3 1 2 3 1 2 3
=
3 2 1 2 1 3 2 3 1
Notice that order matters:
1 2 3 1 2 3 1 2 3
=
2 1 3 3 2 1 3 1 2
Clearly S is ab elian, since it consits of only the identity element.
1
We'll see that S is ab elian in fact it's cyclic since it has only two
2
elements. However, wehave seen that S is not ab elian and in general:
3
THEOREM 2 If n 3 then S is non-ab elian. n
5.1 Symmetric Groups 3
PROOF Let and b e the following two elements in S :
n
1 2 3 4 ::: n 1 2 3 4 ::: n
= = :
3 2 1 4 ::: n 2 1 3 4 ::: n
Then
1 2 3 4 ::: n 1 2 3 4 ::: n
= = :
3 1 2 4 ::: n 2 3 1 4 ::: n
EXAMPLE 2 Find the following inverse:
1
1 2 3 1 2 3
=
2 3 1 3 1 2
THEOREM 3 jS j = n!:
n
PROOF jS j is just the numberofways the integers 1 through n can arranged.
n
In other words in how many di erentways can we ll in the blanks:
1 2 ::: n
:::
Well wehave n choices for the rst entry, and then n 1choices for the
next entry, and so on yielding a total of
n n 1 1= n!
total choices.
For example jS j =3!=6aswehave already seen. jS j = 2!=2.
3 2
If we write out the Cayley table for S we see that it is ab elian. Note
2
that the square of every element is the identity.
1 2 1 2
1 2 2 1
1 2 1 2 1 2
1 2 1 2 2 1
1 2 1 2 1 2
2 1 2 1 1 2
EXAMPLE 3 Compare the symmetries of an equilateral triangle to S . Write each
3
elementinD as a p ermutation.
3
EXAMPLE 4 Compare D to S . Are they the same group?
4 4
4 Math 375
Cycle Notation
It will provemuch more convenient to use a di ernttyp e of notation to
denote the individual p ermutations of S . This cycle notation is useful
n
b ecause it reveals the structure of individual elements in the group, much
as p owers of a generator indicate the nature of a cyclic group. Here's
howitworks.
Consider the following elements of S :
5
1 2 3 4 5 1 2 3 4 5
= =
2 3 1 5 4 4 3 5 1 2
Notice that breaks quite naturally into two parts:
1 ! 2 ! 3 ! 1 and 4 ! 5 ! 4
Each of these pieces is called a cycle. The rst is a three-cycle b ecause
it contains three di erent elements, the second is a two-cycle. Eachof
these cycles can b e represented more compactly without the arrows:
1; 2; 3 and 4; 5
Let's lo ok at . This time the cycles are:
1 ! 4 ! 1 and 2 ! 3 ! 5 ! 2
or
1; 4 and 2; 3; 5
When using cycles we adopt the convention that if an elementis
missing from the cycle, then it is mapp ed to itself.
EXAMPLE 5 What p ermutation do es the cycle 2; 4; 3 representin S ?
6
1 2 3 4 5 6
2; 4; 3 =
1 4 2 3 5 6
EXAMPLE 6 Let = 1; 3; 6 2; 4; 5 and = 1; 2; 3 4; 5 in S . Then =
6
4; 2; 6; 1 5 3 = 4; 2; 6; 1 . Do some more.
EXAMPLE 7 What is the inverse of a single cycle: = 4; 2; 3; 5 ? This is easy to
sp ot in matrix form. If
1 2 3 4 5
= 4; 2; 3; 5 = ;
1 3 5 2 4
then
1 2 3 4 5
1
= = 5; 3; 2; 4 :
1 4 2 5 3 More generally:
5.1 Symmetric Groups 5
THEOREM 4 The inverse of a cycle is the cycle in inverse reverse order. Further
the inverse of a pro duct of cycles is the pro duct of the inverse cycles in
reverse order.
REMARK The second part of the theorem is the usual statement ab out inverses of
pro ducts b eing the pro duct of the inverses in reverse order.
DEFINITION 5 Two cycles are called disjoint if they contain no term in common.
For example 1; 2; 6 and 3; 4 are disjoint but 1; 2; 4 and 3; 4; 5
are not. One of the basic facts ab out p ermutations is that
THEOREM 6 Any p ermutation can b e written as a pro duct of disjoint cycles.
The details of the pro of are presented in the text, but note that
this is howwe rst got started on lo oking at cycles. Given an individual
example of a p ermutation it is easy to see how to split it into disjoint
cycles. Consider
1 2 3 4 5 6 7 8 9
:
5 9 3 8 6 1 4 7 2
We simply start the rst cycle with 1, continue backuntil we get 1.
Then start the second cycle with the smallest remaining unused num-
ber until we get back to it and so on. In this case the cycles are:
1; 5; 6 2; 9 3 4; 8; 7 :
THEOREM 7 Disjoint cycles commute.
PROOF Let the disjoint cycles b e = a ;a ;:::;a and = b ;b ;:::; ,
1 2 m 1 2 n
where and havenoentries in common. Let the full set S b e given
by
S = fa ;a ;:::;a ;b ;b ;:::; ;c ;:::;c g:
1 2 m 1 2 n 1 k
Wewant to prove that = .To do this, wemust show that a =
x for all x 2 S . There are three cases: x is either an a, b,or c.
Supp ose that x = a . Then
i
x = a = a = a = a :
i i i+1 mo d m i+1 mo d m
while
x = a = a = a = a :
i i i i+1 mo d m
Hence the p ermutations agree on the a's. A similar argumentworks for
the b's. The c's are the easiest of all. If x = c , then
i
x = c = c = c = c
i i i i
while
x = c = c = c = c :
i i i i
Again the p ermutations are the same.
6 Math 375
5.2 More ab out S
n
The Order of a k -cycle
EXAMPLE 8 Consider the following elements of S : = 2; 3 and = 1; 2; 4 , and
5
= 1; 3; 4; 2 Find the orders of each of these elements.
2
Taking rst, = 1; 4; 2 , every entry is moved two places down
the cycle. So if wehaveak -cycle, moving eachentry k places down the
th
cycle takes the entry back to its starting p oint, i.e., the k power of
a k -cycle yields the identity and no smaller p ower will. It is clear that
2 3 4
= e, = e, and = e. In fact, this proves the general theorem:
THEOREM 8 If is a k -cycle in S , then j j = k .
n
THEOREM 9 The order of the pro duct of disjoint cycles is the least common mutliple
of their lengths orders .
PROOF Asume that j j = m and j j = n. Let k = lcm m; n . Then it follows
k k
that = e = since j jj k and j jj k . Since the cycles are disjoint,
k k k
they commute, so = = ee = e. Consequently j jj k .
But is k the smallest p ositivepower with this prop erty? Supp ose
t t t t t
instead that it were t. Then = = e implies that = .
t t
But and are disjoint. The only waytwo disjoint cycles can b e
t t
the same is if b oth are empty. That is, = = e rememb er xed
symb ols are not included in the cycles. But now it follows that b oth
m and n divide t. That is, t is a common multiple of m and n so
k = lcm m; n t. Therefore, k = t.
EXAMPLE 9 Give some. If the cycles are not disjoint, simply rewrite them as disjoint,
and then apply the theorem.
Disjoint cycles are esp ecially helpful for order calculations as wehave
seen. But p ermutations can b e written as a di erent sort of pro duct as
the follwing discussion shows.
Two Cycles or Transp ositions
The simplest sort of p ermutation is the one that shues two elements,
that is, a two-cycle. In fact two-cycles are the building blo cks of the
entire p ermutation group. Two-cycles are also called transp ositions.
5.2 More ab out S 7
n
EXAMPLE 10 Consider the p ermutation 1; 2; 3 . It can b e written as a pro duct of
two-cycles:
1; 2; 3 = 1; 3 1; 2 :
Notice that it can b e written as a di erent pro duct of two-cycles:
1; 2; 3 = 2; 3 1; 3 :
In fact, this is a particular example of
THEOREM 10 Any k -cycle in S can b e written as a pro duct of transp ositions two-
n
cycles . Here n>1 or else wehave S = feg.
1
PROOF If wehave a 1-cycle, then it is the identity element which can b e written
2
as 1; 2 = ; 2 1; 2 = e.Nowifwehavea k -cycle were k 2 then we
can work out the pro duct just as we did in the example ab ove:
a ;a ;:::;a = a ;a a ;a ::: a ;a :
1 2 k 1 2 1 3 1 k
EXAMPLE 11 5; 3; 1; 2 = 5; 2 5; 1 5; 3
COROLLARY 11 Any elementofS can b e written as a pro duct of transp ositions.
n
PROOF Any elementinS can b e written as a pro duct of disjoint cycles, eachof
n
which can b e written as a pro duct of two-cycles. So every element can
b e broken down eventually into a pro duct of transp ositions.
EXAMPLE 12
1 2 3 4 5 6
= 1; 2; 5; 3 4; 6 = 1; 3 1; 5 1; 2 4; 6 :
2 5 1 6 3 4
It is also the case that
1 2 3 4 5 6
= 4; 6 2; 5; 3; 1 = 4; 6 2; 1 2; 3 2; 5 :
2 5 1 6 3 4
But notice that even though the numb er of transp ositions in each
decomp osition is di erent, in b oth cases the numb er of transp ositions is
even.
DEFINITION 12 A p ermutation is even if it can b e written as the pro duct of an even
numb er of p ermuatations. It is odd if it can b e written as a pro duct of
an o dd numb er of transp ositions.
For this distinction to b e useful, it is necessary that no o dd p ermu-
tation b e even. That is, either a p ermutation should always decomp ose
into an o dd numb er of transp ositions or always into an even number
even though the actual numb er of transp ositions in the decomp osition
can vary . In fact, this is the case:
8 Math 375
THEOREM 13 No p ermuatation is b oth o dd and even.
PROOF The pro of is in the text; read through it. The main idea is that the
identity p ermutation can only b e written as an even pro duct of transp o-
sitions. The general result easily follows from this. For if a p ermutation
was b oth even and o dd, then its inverse would b e b oth even and o dd.
Take the p ermutation in its even representation and multiply by its in-
verse in its odd representation. The reuslt is pro duct with an o dd number
of transp ositions, but the result is also the identity which is supp osed to
be even.
COROLLARY 14 A k -cycle is even if k is o dd; a k -cycle is o dd if k is even.
PROOF This follows immediately from the the metho d we use to break a k -cycle
into transp ositions. A k -cycle can always b e written as the pro duct of
k 1 transp ositions see the factors in Examples 17.3 and 4 .
EXAMPLE 13 Let A = the set of even p ermutations in S . Show that A is a subgroup
n n n
of S . A is called the alternating group of degree n.
n n
CLOSURE The pro duct of twoeven p ermutations will again haveaneven number
of factors, so A is closed.
n
1
INVERSES Notice that ab = ab . Thus the inverse of a pro duct of even
p ermutation using the general formula for the inverse of a pro duct is
the pro duct of the same transp ositions in reverse order, hence is even
again.
EXAMPLE 14 Do the o dd p ermutations form a subgroup of S ? n
5.2 More ab out S 9
n
EXAMPLE 15 We can easily write out the elements of A and A b ecause we know that
3 4
the length of a cycle determines whether it is o dd or even.
Here are the 6 elements of S , let's pick out the elements of A :
3 3
1 2 3 1 2 3
= e 2 A = 2 3 2= A
3 3
1 2 3 1 3 2
1 2 3 1 2 3
= 1 2 3 2 A = 1 2 2= A
3 3
2 3 1 2 1 3
1 2 3 1 2 3
= 1 3 2 2 A = 1 3 2= A
3 3
3 1 2 3 2 1
For S and A the situation is similar. S has 24 elements:
4 4 4
1 2 3 4 1 2 3 4
= 2 3 2= A = e 2 A
4 4
1 3 2 4 1 2 3 4
1 2 3 4 1 2 3 4
= 1 2 2= A = 1 2 3 2 A
4 4
2 1 3 4 2 3 1 4
1 2 3 4 1 2 3 4
= 1 3 2 2 A = 1 3 2= A
4 4
3 1 2 4 3 2 1 4
1 2 3 4 1 2 3 4
= 3 4 2= A = 2 3 4 2 A
4 4
1 2 4 3 1 3 4 2
1 2 3 4 1 2 3 4
= 1 2 3 4 2 A = 1 2 3 4 2= A
4 4
2 1 4 3 2 3 4 1
1 2 3 4 1 2 3 4
= 1 3 4 2 A = 1 3 4 2 62 A
4 4
3 2 4 1 3 1 4 2
1 2 3 4 1 2 3 4
= 2 4 3 2 A = 2 4 2= A
4 4
1 4 2 3 1 4 3 2
1 2 3 4 1 2 3 4
= 1 2 4 2 A = 1 2 4 3 2= A
4 4
2 4 3 1 2 4 1 3
1 2 3 4 1 2 3 4
= 1 3 2 4 2= A = 1 3 2 4 2 A
4 4
3 4 2 1 3 4 1 2
1 2 3 4 1 2 3 4
= 1 4 2 2 A = 1 4 3 2 2= A
4 4
4 1 3 2 4 1 2 3
1 2 3 4 1 2 3 4
= 1 4 3 2 A = 1 4 2= A
4 4
4 2 1 3 4 2 3 1
1 2 3 4 1 2 3 4
= 1 4 2 3 2 A = 1 4 2 3 2= A
4 4
4 3 2 1 4 3 1 2
There are two imp ortant observations. First, we see that S is a
3
subset and hence a subgroup of S just take a lo ok at the rst elements
4
of S . In general we can always view S as a subgroup of S if m 4 m n Second, half the elements in S are even; the same is true for S . In fact, 3 4 THEOREM 15 jA j = jS j=2= n!=2. n n 10 Math 375 PROOF We need to show that half the elements of S are even. Let be any n two-cycle of S say = 1 2 . Consider the mapping f : S ! S by n n n f = . Notice that f is b oth surjective and injective. SURJECTIVE Let 2 S .Wemust nd 2 S so that f = . But n n 1 f = = = : INJECTIVE = = = . 1 2 1 2 1 2 Notice that f maps the even p ermutations to the o dd ones and vice versa since it adds one transp osition to the p ermutation in a one-to- one, onto fashion. Hence there must b e the same numberofeven and o dd p ermutations, that is half of S is even. n Some Examples EXAMPLE 16 Do Gallian page 110 50. SOLUTION If we let denote the shue in question, then we are given that 1 2 3 4 5 6 7 8 9 10 J Q K 2 = 10 9 Q 8 K 3 4 A 5 j 6 2 7 = 1 10 J 6 3 Q 2 9 5 13 7 4 8 : 2 Notice that must have b een a 13-cycle since is. If we expressed the original as a pro duct of disjoint cycles, p owers of simply p ermute the elements of those cycles among themselves. Thus, j j = 13. So 2 7 14 13 = = = : But 2 7 = 1 9 10 5 J K 6 7 3 4 Q 8 2 : EXAMPLE 17 Do Gallian page 109 28. 99 1 SOLUTION = 123 145 = 14523 , so = = 32541 : EXAMPLE 18 Do Gallian page 109 37. k SOLUTION Let b e a 10-cycle. Do you see that there are p owers so that is not k a 10-cycle? Those p owers share a divisor with 10. So is a 10-cycle i k gcd 10;k = 1 < >=< >. 5.3 Dihedral Groups and Symmetric Groups 11 5.3 Dihedral Groups and Symmetric Groups S D = 3 3 Let us now reconsider thte rigid motions of an equilateral triangle with vertices lab elled 1, 2, 3. Then every motion of the the triangle can b e thoughtofasapermutation in S . Sp eci cally,we can set up the 3 following corresp ondence: r ! 1 r ! 1; 2; 3 r ! 1; 3; 2 0 120 240 0 v ! 2; 3 d ! 1; 3 d ! 1; 2 : If we denote this corresp ondence or mapping by , then we see that : D ! S is one-to-one and onto. Further, one can check that 3 3 resp ects the group op erations involved. That is: if x; y 2 D , then 3 xy = x y : Notice the group op eration on the left takes place in D and the op eration 3 on the right takes place in S .For example: 3 0 vr = d = 1; 2 and a r = 2; 3 1; 3; 2 = 1; 2 : 240 240 This is an example of an group isomorphism, that is, a one-to-one, onto map b etween groups that resp ects the group op erations. We will study such maps in great detail later. Notice that the elements of D can b e listed in another way.Ifwe 3 let = r then the following six elements are distinct: 120 2 2 fe; ; ; h; h; h g D : 3 But since jD j = 6 these six elements must comprise all of D . In other 3 3 words D , while it is not cyclic, it is generated by two elements. 3 D and S 4 4 We can construct a similar corresp ondence b etween D and some of the 4 elements of S . Lab el the vertices of a square as 1, 2, 3, 4. As with the 4 motions of a triangle, the particualr righid motion in D is determined 4 12 Math 375 by whether the square is face up or face down and by where the vertex 1 ends up. Hence jD j =2 4 = 8. this time the rotations are 4 r = 1 r = 1; 2; 3; 4 r = 1; 3 2; 4 r = 1; 4; 3; 2 0 90 180 270 0 h = 1; 4 2; 3 v = 1; 2 3; 4 d = 1; 3 d = 2; 4 : Let = r .Now jD j = 8 and the following 8 elements are distinct 90 4 and in D : 4 2 3 2 3 fe; ; ; ; h; h; h ;h g D : 4 Therefore this must b e the entire group. We knownowtwo di erent subgroups of S , namely A and D . 4 4 4 Are there others? All the cyclic groups, S , and there are others see 3 text . This same pro cess that wehave carried out with D and S can 4 4 b e done with D and S , where D represents the motions of a regular n n n n-sided p olygon. This lime let 1 2 3 4 ::: n 1 n = 1; 2;:::;n and h = 1 n n 1 n 2 ::: 3 2 360 and h represents the re ection across Here represents a rotation of n the line through vertex 1. You should b e able to show that 2 n 1 2 n 1 D = fe; ; ;:::; ; h; h; h ;:::;h g: n That is D is generated bytwo elements: its smallest rotation and any n ip. See the optional exercise on Problem Set 11.