Math 301 Assignment 6 solutions

1 a Let 1 denote the donothing symmetry

Let r denote a rotation of 3605 720 inthe clockwise Cw direction

Then it denotes a rotation of 1440 CWand r 2160 Cwand r 288 Cw

For each kite 5 fi denotes the corresponding reflection acrossthe axis drawn in the diagramThere are 10 symmetries in all

Ds l r R2 P r f fz fz fy f5Each symmetry could also be representedby the permutation itinduces on the labels on the vertices

I E f 25 34r vs 12345 fz l 2 35r2 13524 f 13 45P I 4253 fy 14 23TY I 5432 fg 15 24

b c Cayley table

1 r r r r f f f f

foli r r2 r's r't f f f fo fr r r r r l f f f f for2 r2 r f't I r fr r r l r r ft EI Ifs tf ftp't r't I r r p 3

f f f f fo f s I r R2 P pmfz fz fz fu f5 f r l r r p'sf f fo f f f z r r l r r2fy f Es f f f r p r't I rft f f f f fo r r r r l

Do is net abelian since for example rf f r

d Since IDs1 10 it can only have subgroups of orders1,2 5 10 The subgroup of order 1 is 13 and thesubgroup of order 10 is Ds itself

Subgroups of order 2 There are 5 of them eachcorresponding to an element of order 2

Lf fz Lf fy FsSubgroups of order 5 First observethat by Corollary 11.1every non identity element in a subgroup of order 5must have order 5 too This means none of thereflections can be in a subgroup of order 5 Theonly subgroup of order 5 is

r I r r r T

2 Dn is not cyclic To see why we will provide a fewdifferent arguments

Reason 4 There is no element of order 2h The rotationshave order at most n sincethey form a

subgroup of size n and the reflections eachhave order 2

Reason 2 i Dn is not abelianForexample consider reflechin f as drawn left

i and rotation r Thenn l f r takes vertex 1 to 2

i 2 butf rf takes vertex 1 to n

Thereforf r rfi

since Dn is not abelian it cannot be cycledReason3 Dn has at least n elements of order 2 namely

the reflections However a cyclicgroup canhave at most 1 element of order 2Theorem 11.7

3 By Theorem 11.9 a the generators of Zeo are 1 3,39

4 By Theorem 11,9 c the elements of order 8 areK 1,000,000 where gcd K 8 l

There are 4 elements1,000,000 3,000,000 5,000,000 7,000,000

5 We'll use Theorem 11.9 Cb to find the subgroups andfor each subgroup we use Theorem 11.9 a to getall generators

a 71,2 The subgroups of Liz are of sizes 1,2 3,4 6,12

order 12 subgroup 2,2 Li 5 7 L 11order 6 subgroup L2 92,4 6,8 10 toorder 4 subgroup 3 0,36,93 9order 3 subgroup 24 o 4,8 8order 2 subgroup LG 0,6order 1 subgroup so o

We can express this in the following diagram graph12 order 12

Iorder6 L2 0,24,68,103

23 0,3 6,9 29 order4

order3 44 04,83 287

Y0,63 order 2

Lo O order 1

b In 7 if the order of a subgroup divides 17 so it iseither 17 i.e the whole group or 1 i.e the trivialsubgroup

7 iz order 17

Io 03 order I

6 a L 1234 E Ll 234 Cl3 24 C 1432 L Syhas order 4

b Dy e Sc of order 8 where we view the elements of Dcas permutations of the vertices of a square under the8 symmetries

7 G has 0683 4 elements of order 8 by Theorem 11.7 If ais one element of order 8 then

a as as a7

are all the elements of order 8 Note I 3,5 7 are

precisely the numbers less than and relatively prime to 8

8 If 161 33 then the possible orders of elements areI 3 11 and 33

Towards a contradiction suppose G does not have an elementof order3 Then it also can't have an element of order 33if ghas order 33 then g has order3 Therefore underthisassumption the elements of G have order either 1 or 11There is only one element of order 1 namely the identity Therewould then need to be 32 elements of order 11 We'll showthat this is impossible

First I'll give a quick argument as to why this is impossibleAfterwards I'll give a more constructive argumentConsider

x'EG l x e and x t eSince every element in G is assumed to have order11 or 1then this set is G Ee and must have 32 elements Howeverfor each element b of order 11 it has 10 related elements

b b b3 b ball of which have order 11 Therefore can be splitup into disjoint sets of size 10 which means its cardinalityis divisible by 10 But 101132 so we have a contradiction

We've seen this argument twice now HW 3 question4

The following is a more constructive argument showing that allnon identity elements cannothave order11

Consider an element of order 11 say a Thena e a arr ooo a

Pick another element in G that is not in La call this b Thenb has order 11 and

b e b b b

Claim as ncb e

Pf If x C La ncb sit X e then x hasorder11 so La L c bD

Now La n Cb consists of 21 elements of GLet CE G La u Cbs and by a similar argument

a b Lc

only share the identity element Together they consist of31 elements of G So there is still another element

de G La ucb ucc

which has order 11 Butd e d d2 d

would consist of 10 elements in G which are not aira u Lb ucc This is impossible since it would imply161 241

Therefore G must have an element of order 3D

9 Since the order of an element divides the order ofthe group then any nonidentity element most haveorder p Since 161 p then any nonidentity elementgenerates the group Therefore G is cyclic

10 Let 6 be a group with the property that the square ofevery element is the identity Let g hEG be anytwo elements Then

gh e by the propertyof thisgroupghgh e

ghg high n ggh hg since h2 e gZ then h eh g Ig

Therefore 6 is abelian

11,12 This is done in Section 151.4 results are summarizedin Theorem 15 1.3

oh DIEi

neverAn if NZ 6 n odd

13 To determine which moves are possible we first construct theRubik's cube group RG in SageMath as a subgroup of Syoogenerated by R L U D F B

a Due to symmetry of the cube it suffices to check thisfor any corner so I'll check whether it is possible totwist the ufl cubie in the counterclockwise directionThis is equivalent to determining whether the 3 cycle6 11 17 is in Rcs

Sagemath tells us it is not in RG so a CCW twist ofa corner is impossible Moreover a Cw twist is impossibletoo since its square would be a Ccw twist

b First we check if it is possible to twist the ufl and cefr coble'sin the same direction say clockwise Cw

Therefore it is mpo Similarly twisting two corners Ccwwould be impossible since it is the inverse of a Cw twistBy symmetry we see it is impossible to twist any two adjacentcorners in the same direction

What about two corners in general No it is notpossible Ifit were then we could do two adjacent ones via conjugationwhich is impossible

c We'll check if it is possible to twist nfl cow and Ufr Cw

Therefore it is pork