Mathematics 343a solutions
1.2.1 (a) (124)(14)(23) = (1)(234) = (24)(23); parity even + even = even = odd + odd. For
inverses, we note that [(124)(14)(23)]−1 = (234)−1 = (243) = [(14)(23)]−1 (124)−1 .
(b) [(146)(27)(35)]6 = (146)6 (27)6 (35)6 (since the cycles are disjoint, they commute) and we
obtain e as the result; parity of the given permutation is even + odd + odd = even.
(c) (234)(12345) = (13245) = (15)(14)(12)(13); parity even + even = even = odd + odd
+ odd + odd. For inverses, [(234)(12345)]−1 = (13245)−1 = (15423) = (15432)(243) =
(12345)−1 (234)−1 .
(d) (362)(15)(42) = (15)(2436) = (15)(26)(23)(24). The permutation is even; the inverse is
(15)(2634).
1.2.2 (123)(456) = (142536)2 . A general result is: if i · j = n, and σ is a product of i disjoint
cycles of length j , these cycles can be “interleaved” to write σ as a single cycle of length n
raised to the power i.
1.2.3 First write the permutation as a product of (an even number of) transpositions. Grouping
these in pairs, we find two cases:
(a) (ij)(ik) = (ijk);
(b) (ij)(kl) = (ikl)(ikj).
In either case we may replace each pair of transpositions by one or more 3-cycles.
1.5.1 Suppose we have two inverses b and c for an element a of a group. Then we may write
b = b · e = b · (a · c) = (b · a) · c = e · c = c, using the properties of identity, inverses, and
associativity.
1.5.2 The statement is equivalent to g = g −1 . We pair each non-identity element with its inverse.
Since |G| is even there are an odd number of non-identity elements, and so the above pairing
must result in at least one element paired with itself.
1.5.3 Suppose |G| = p (a prime) and g is a non-identity element of G. The order of the subgroup
hgi must divide p; since g 6= e we find this order is p, and so hgi = G. The converse is similar
since the order of the subgroup generated by any non-identity element must be n. In general,
the number of generators of Cn is φ(n), the number of integers between 1 and n relatively
prime to n.
1.5.4 The left cosets of H = {e, g, g 2 , g 3 } are H itself and hH = {h, g 3 h, g 2 h, gh}. The right
cosets of H are H and Hh = {h, gh, g 2 h, g 3 h}.
20
Another subgroup of D4 of order four is K = {e, g 2 , h, g 2 h} (note that this subgroup is
essentially different from the first since it is not cyclic; how many others can you find?) The
left cosets are K and gK = {g, g 3 , gh, g 3 h}; the right cosets are K and Kg = {g, g 3 , g 3 h, gh}.
Note that in both cases we found the set of left cosets to be the same as the set of right cosets.
This is not true in general in non-commutative groups like D4 —can you find an explanation?
1.5.5 Every permutation is either even or odd. Furthermore the coset An (12) contains all odd
permutations. To see this suppose σ is odd; then π = σ ◦ (12) is even, so π ∈ An . Thus
π ◦ (12) is in An (12); but π ◦ (12) = σ since (12)2 = e. Since |An (12)| = |An | we are done.
1.5.6 (a) D5 is an easy example of a subgroup of order 10 in S5 .
(b) D7 has order 14; 3 is not a factor of 14 so there is no such subgroup.
(c) If H has order 5 it is cyclic (H ' C5 ). Thus we must find an even permutation σ of
{1, 2, 3, 4, 5, 6} so that σ 5 = e, and σ i 6= e for 0 < i < 5. One such permutation is (12345).
2.3.1 (a)
cycle structure
e = (1)
g = (1234)
g 2 = (13)(24)
number, s
1
2
1
|Fix σ|
s · |Fix σ|
34
= 81
3
2
3 =9
81
6
9
4
96
The number of colourings is (81 + 6 + 9)/4 = 24.
(b)
cycle structure
number, s
|Fix σ|
s · |Fix σ|
e = (1)
g = (1234)
g 2 = (13)(24)
h = (24)
1
2
3
2
81
3
9
3
3 = 27
81
6
27
54
8
168
The number of colourings is 168/8 = 21. You should find three-colourings of the vertices
of the square which are equivalent if reflections are allowed, but are not equivalent if only
rotations are allowed.
2.3.2 (a) The appropriate group is D6 .
21
cycle structure
number, s
e = (1)
g = (123456)
g 2 = (135)(246)
g 3 = (14)(25)(36)
h = (26)(35)
hg = (12)(36)(45)
1
2
2
1
3
3
|Fix σ|
¡6¢
3
s · |Fix σ|
= 20
0
2
0
4
0
12
20
0
4
0
12
0
36
The number of necklaces is 36/12 = 3.
(b) Here we use D9 .
cycle structure
e = (1)
g = (123456789)
g 3 = (147)(258)(369)
h = (29)(38)(47)(56)
number, s
1
6
2
9
|Fix σ|
¡9¢¡6¢
= 1680
0
3! = 6
0
3
3
18
s · |Fix σ|
1680
0
12
0
1692
The number of necklaces is 1692/18 = 94.
(c) The group is D8 .
cycle structure
e = (1)
g = (12345678)
g 2 = (1357)(2468)
g 4 = (15)(26)(37)(48)
h = (28)(37)(46)
hg = (12)(38)(47)(56)
number, s
1
4
2
1
4
4
|Fix σ|
¡8¢¡5¢
= 560
0
0
0
2 · 3 · 2 = 12
0
3
3
16
s · |Fix σ|
560
0
0
0
48
0
608
The number of necklaces is 608/16 = 38.
(d) D10 :
22
cycle structure
number, s
|Fix σ|
1
4
4
1
5
5
210 = 1024
21 = 2
22 = 4
25 = 32
26 = 64
25 = 32
e
g = (123456789 10)
g 2 = (13579)(2468 10)
g 5 = (16)(27)(38)(49)(5 10)
h = (2 10)(39)(48)(57)
hg = (12)(3 10)(49)(58)(67)
20
s · |Fix σ|
1024
8
16
32
320
160
1560
The number of necklaces is 1560/20 = 78.
2.3.3 This problem is essentially asking for the number of 3-colourings of the hexagon under D6 :
cycle structure
number, s
e = (1)
g = (123456)
g 2 = (135)(246)
g 3 = (14)(25)(36)
h = (26)(35)
hg = (12)(36)(45)
|Fix σ|
s · |Fix σ|
36
1
2
2
1
3
3
= 729
31
32
33
34 = 81
33 = 27
12
729
6
18
27
243
81
1104
The number of compounds is 1104/12 = 92.
2.3.4 There are three different kinds of symmetry of the tetrahedron: the identity, rotations keeping
one vertex fixed (eight of these), and rotations about an axis bisecting two opposite sides
(three of these). We count the 2-colourings—the work for 3 or n is essentially the same.
cycle structure
e
(123)
(12)(34)
number, s
1
8
3
|Fix σ|
24
22
s · |Fix σ|
= 16
=4
22
16
32
12
12
60
There are 60/12 = 5 different 2-colourings.
2.3.5 To prove that g is 1–1 on X, suppose that g(x1 ) = g(x2 ). Then g −1 g(x1 ) = g −1 g(x2 ) and so
x1 = x2 . Furthermore, g is onto since for any x ∈ X, g −1 (x) is in X and gg −1 (x) = x. Note
that this proof uses only the definition of group action and does not require either the group
or the set to be finite.
23
2.3.6 As found before, there are 24 symmetries of the cube: the identity e, rotations about an axis
through the centres of two opposite faces (there are two kinds here: g, rotation by π/2, and
g 2 , by π), rotations h by π about an axis through the centres of diagonally opposite sides, and
rotations f by 2π/3 and 4π/3 about the axis through diagonally opposite vertices. Counting
the 2-colourings, using the labeling of the cube as above,
cycle structure
number, s
e
g = (1243)(5687)
g 2 = (14)(23)(58)(67)
h = (12)(36)(45)(78)
f = (1)(235)(476)(8)
|Fix σ|
s · |Fix σ|
28
1
6
3
6
8
= 256
22 = 4
24 = 16
24 = 16
24 = 16
24
256
24
48
96
128
552
The number of 2-colourings is 552/24 = 23. Similar work gives the number of n-colourings
to be (n8 + 17n4 + 6n2 )/24. Note that g 2 and h above have the same cycle structure and so
could be grouped together (though they are different from a geometric point of view).
3.1.1 We tabulate the work as in a previous example, except that here we require the coefficient
on c31 c22 c3 :
σ∈G
s
polynomial
coefficient of c31 c22 c3
(a)(b)(c)(d)(e)(f )
(a)(bcde)(f )
(a)(bd)(ce)(f )
(ab)(ce)(df )
(abc)(def )
1
6
3
6
8
(c1 + c2 + c3 )6
6(c1 + c2 + c3 )2 (c41 + c42 + c43 )
3(c1 + c2 + c3 )2 (c21 + c22 + c23 )2
6(c21 + c22 + c23 )3
8(c31 + c32 + c33 )2
0
3 · 2 · 2 = 12
0
0
24
¡6¢ ¡3¢
3
2 = 60
72
There are 72/24 = 3 such colourings (you can check this directly).
3.1.2 Using previous work where possible,
2.3.1 (a) The cycle index of C4 can be found from the table on p. 21 to be PC4 = (x41 +
x22 + 2x4 )/4. For the total number of 3-colourings, set x1 = x2 = x3 = 3 to obtain
(34 + 32 + 2 · 3)/4 = 24 3-colourings.
(b) For D4 the cycle index is (x41 +2x21 x2 +3x22 +2x4 )/8; we find (34 +2·32 ·3+3·32 +2·3)/8 =
21 colourings.
24
2.3.2 (a) The appropriate group is D6 .
cycle structure
s
polynomial
e = (1) · · · (6)
g = (123456)
g 2 = (135)(246)
g 3 = (14)(25)(36)
h = (1)(26)(35)(4)
hg = (12)(36)(45)
1
2
2
1
3
3
(c1 + c2 )6
2(c61 + c62 )
2(c31 + c32 )2
(c21 + c22 )3
3(c1 + c2 )2 (c21 + c22 )2
3(c21 + c22 )3
coefficient of c31 c32
¡6¢
= 20
0
¡2¢
2· 1 =4
0
3 · 2 · 2 = 12
0
12
The number of necklaces is 36/12 = 3.
(b) Here we use D9 .
3
36
cycle structure
s
polynomial
e = (1) · · · (9)
g = (123456789)
g 3 = (147)(258)(369)
h = (1)(29)(38)(47)(56)
1
6
2
9
(c1 + c2 + c3 )9
6(c91 + c92 + c93 )
2(c31 + c32 + c33 )3
9(c1 + c2 + c3 )(c21 + c22 + c23 )4
coefficient of c31 c32 c33
¡9¢¡6¢
= 1680
0
2 · 3 · 2 = 12
0
3
3
18
The number of necklaces is 1692/18 = 94.
(c) The group is D8 .
1692
cycle structure
s
polynomial
e = (1) · · · (8)
g = (12345678)
g 2 = (1357)(2468)
g 4 = (15)(26)(37)(48)
h = (1)(28)(37)(46)(5)
hg = (12)(38)(47)(56)
1
4
2
1
4
4
(c1 + c2 + c3 )8
4(c81 + c82 + c83 )
2(c41 + c42 + c43 )
(c21 + c22 + c23 )4
4(c1 + c2 + c3 )2 (c21 + c22 + c23 )3
4(c21 + c22 + c23 )4
coeffiecient of c31 c32 c23
¡8¢¡5¢
= 560
0
0
0
4 · 2 · 3 · 2 = 48
0
3
3
16
608
The number of necklaces is 608/16 = 38.
(d) D10 : here we require the complete cycle index, because any number of each colour
may be used.
25
cycle structure
s
cycle structure terms
e
g = (123456789 10)
g 2 = (13579)(2468 10)
g 5 = (16)(27)(38)(49)(5 10)
h = (1)(2 10)(39)(48)(57)(6)
hg = (12)(3 10)(49)(58)(67)
1
4
4
1
5
5
x10
1
4x10
4x25
x52
5x21 x42
5x52
20
5
2
2 4
(x10
1 + 5x1 x2 + 6x2 + 4x5 + 4x10 )/20
The cycle index is thus
we find the number of necklaces is 1560/20 = 78.
and setting x1 = x2 = · · · = 2
2.3.3 As in the previous problem we require the complete cycle index of the group, which in
this case is D6 .
cycle structure
s
cycle structure terms
e = (1)
g = (123456)
g 2 = (135)(246)
g 3 = (14)(25)(36)
h = (1)(26)(35)(4)
hg = (12)(36)(45)
x61
2x6
2x23
x32
3x21 x22
3x32
1
2
2
1
3
3
12
(x61
The cycle index is
+ 3x21 x22 + 4x32 + 2x23 + 2x6 )/12; with x1 = x2 = · · · = 3 we obtain
1104/12 = 92 distinct compounds.
2.3.4
cycle structure
s
cycle terms
e
(123)
(12)(34)
1
8
3
x41
8x1 x3
3x22
12
The cycle index is
2-colourings.
(x41
+ 8x1 x3 + 3x22 )/12. Substituting x1 = · · · = 2 we have 60/12 = 5
26
2.3.5
cycle structure
s
cycle terms
e
g = (1243)(5687)
g 2 = (14)(23)(58)(67)
h = (12)(36)(45)(78)
f = (1)(235)(476)(8)
1
6
3
6
8
x81
6x24
3x42
6x42
8x21 x23
24
(x81
Cycle index:
+
8x21 x23
+ 9x42 + 6x24 )/24. For 2-colourings, we find 552/24 = 23.
3.1.3 The cube has 12 edges, which we will label A, B, . . . , L for convenience. The letters are
matched up with the diagrams on p. 17 as shown in the following table:
endpoints
12
24
34
13
15
26
48
37
56
68
78
57
label
A
B
C
D
E
F
G
H
I
J
K
L
The 24 rotations of the cube produce permutations of {A, . . . , L}:
cycle structure
s
cycle structure terms
e
g = (ABCD)(EF GH)(IJKL)
g 2 = (AC)(BD)(EG)(F H)(IK)(JL)
h = (A)(BE)(CI)(DF )(GL)(HJ)(K)
f = (ADE)(BHI)(CLF )(GKJ)
1
6
3
6
8
x12
1
6x34
3x62
6x21 x52
8x43
24
The cycle index is (x112 + 6x21 x52 + 3x62 + 8x43 + 6x34 )/24. Thus there are (212 + 6 · 27 + 3 · 26 + 8 ·
24 + 6 · 23 )/24 = 218 ways to 2-colour the edges. For m colours we find (m12 + 6m7 + 3m6 +
8m4 + 6m3 )/24 colourings.
For the number of colourings with six edges red and six edges blue, we require the coefficient
of c61 c62 when (c1 + c2 ) is substituted for x1 , (c21 + c22 ) is substituted for x2 , etc. The resulting
pattern inventory is
´
1 ³
(c1 + c2 )12 + 6(c1 + c2 )2 (c21 + c21 )5 + 3(c21 + c22 )6 + 8(c31 + c32 )4 + 6(c41 + c42 )3 .
24
The coefficient will be
1
24
ÃÃ
!
12
+6
6
ÃÃ !Ã !
2
0
à !à !!
5
2
+
3
2
27
5
2
à !
à !
6
4
+3
+8
+6·0
3
2
!
= 48.
3.1.4 (a) S4 has 24 elements, of five kinds: the identity, transpositions, 3-cycles, 4-cycles, and
products of two disjoint transpositions. These correspond to five kinds of permutation of Z24 :
σ ∈ S4
s
π∈K
cycle terms
e
(12)
(123)
(1234)
(12)(34)
1
6
8
6
3
(1)(2) · · · (G)
(1)(2)(3)(4)(59)(6A)(7B)(8C)(D)(E)(F )(G)
(1)(2)(395)(4A6)(7BD)(8CE)(F )(G)
(1)(2953)(4AD7)(6B)(8CEF )(G)
(1)(23)(4)(59)(6B)(7A)(8C)(D)(EF )(G)
x16
1
8
6x1 x42
8x41 x43
6x21 x2 x34
3x41 x62
24
4 4
3
4 6
2
8 4
(b) The cycle index is PK = (x16
1 + 6x1 x2 + 3x1 x2 + 6x1 x2 x4 + 8x1 x3 )/24; (c) substituting
x1 = · · · = 2 gives the number of non-equivalent functions to be 95616/24 = 3984.
3.1.5 (a) Proceeding as above we label the ordered pairs 1, 2, . . . , G (lexicographically):
pair
label
1, 1 1, 2 1, 3 1, 4 2, 1 2, 2 2, 3 2, 4 3, 1 3, 2 3, 3 3, 4 4, 1 4, 2 4, 3 4, 4
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
G
Then the permutations corresponding to S4 are
σ ∈ S4
s
π∈G
cycle terms
e
(12)
(123)
(1234)
(12)(34)
1
6
8
6
3
(1)(2) · · · (G)
(16)(25)(37)(48)(9A)(B)(C)(DE)(F )(G)
(16B)(279)(35A)(48C)(DEF )(G)
(16BG)(27CD)(389E)(45AF )
(16)(25)(38)(47)(9E)(AD)(BG)(CF )
x16
1
6x41 x62
8x1 x53
6x44
3x82
24
4 6
5
8
4
The cycle index is (x16
1 + 6x1 x2 + 8x1 x3 + 3x2 + 6x4 )/24.
(b) The number of 2-colourings is PG (2, 2, . . .) = 3044. This is the same as the other two
numbers requested. Binary relations on {1, 2, 3, 4} are subsets of X, and so the number of
them is the number of 2-colourings of X; and a (4 × 4) 0,1-matrix corresponds exactly to such
a binary relation—a pair (i, j) is in the relation iff the row i, column j entry is 1.
3.1.6 This problem is similar to the previous two except that Y has only
we label as a, b, c, d, e, f :
28
¡4¢
2
= 6 elements, which
pair
{1, 2}
{1, 3}
{1, 4}
{2, 3}
{2, 4}
{3, 4}
label
a
b
c
d
e
f
Then the table of cycle structures is
σ ∈ S4
s
π∈G
cycle terms
e
(12)
(123)
(1234)
(12)(34)
1
6
8
6
3
(a)(b)(c)(d)(e)(f )
(a)(bd)(ce)(f )
(adb)(cef )
(adf c)(be)
(a)(be)(cd)(f )
x61
6x21 x22
8x23
6x2 x4
3x21 x22
24
The cycle index is (x61 + 9x21 x22 + 6x2 x4 + 8x23 )/24.
(b) The number of 2-colourings is 264/24 = 11; this is the same as the other two numbers
requested. You should check directly by drawing graphs that there are exactly 11 possibilities.
3.1.7 Here we consider the singleton sets {1}, {2}, {3}, {4} (representing loops) as well as the pairs.
edge
{1}
{2}
{3}
{4}
{1, 2}
{1, 3}
{1, 4}
{2, 3}
{2, 4}
{3, 4}
label
a
b
c
d
e
f
g
h
i
j
The table of cycle structures is
σ ∈ S4
s
π∈G
cycle terms
e
(12)
(123)
(1234)
(12)(34)
1
6
8
6
3
(a)(b)(c)(d)(e)(f )(g)(h)(i)(j)
(ab)(c)(d)(e)(f h)(gi)(j)
(abc)(d)(ehf )(gij)
(abcd)(ehjg)(f i)
(ab)(cd)(e)(f i)(gh)(j)
x10
1
4
6x1 x32
8x1 x33
6x2 x24
3x21 x42
24
4 3
2 4
3
2
The cycle index is (x10
1 + 6x1 x2 + 3x1 x2 + 8x1 x3 + 6x2 x4 )/24, and there are are 2160/24 = 90
graphs.
29