Dummit & Foote chapter 2 selected solutions

8/12/2019 Dummit & Foote chapter 2 selected solutions

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Feynman LiangMATH 350

Assignment #73/10/2014

1, 2, 3, 2.3.11

1 LetG be a group of order 3. Prove that G must be cyclic.

LetG be any group such that the order |G| = 3. Assign labels to the distinct elements ofG

G= {1G, a, b}

where 1G G is the identity (identity law for groups). By 2.2 Proposition 2,

1, a1

, . . . , a|a|

are all distinct elements. Since |G|= 3, it must be that |a| 3 otherwise {1, a, a2, a3} Gare four distinct elements in G, contradicting |G|= 3.Case|a| = 1: Clearly|a| = 1 otherwisea1 = a = 1GsoG = {1G, 1G, b} = {1G, b}whichcontradicts |G|= 3 ().Case |a|= 3: If|a|= 3, then a= {a, a2, a3 =1G} are all distinct elements and by closureofG we havea G. Since|a| = 3 = |G|, it follows thata = Gas sets and thus thetwo groups are equal.Case |a|= 2: Consider |b|.

Case |b| = 1: Not possible otherwise b1 = b = 1G soG = {1G, a, 1G}contradicting

|G|= 3 ().Case |b|= 2: Thenb|b| =b2 =bb= 1G =a a= a

2 =a |a|. This yieldsaa = 1G = a= a

1

bb = 1G = b= b1

after multiplication by inverses. Consider the productab, which is in G by closure andthus must be equal to one of{a, b, 1G}. Ifab = 1G, then by right multiplicationa = b

1 =b and G = {a, b, 1G} = {a, a, 1G} = {a, 1G} contradicting |G| = 3. Ifab = a, then byleft cancellationb = 1Gand|G| = 3 is contradicted. Ifab = b, then by right cancellationa=

1Gand|G|=

3 is again contradicted. All cases lead to contradiction (

).Case|b| = 3: Thenb = {b, b2, b3 = 1G}are all distinct and by similar argument to|a|= 3 we have b= G.

All together, we have that the only two possible cases which do not result in contra-diction are G = a orG = b. Therefore,G must be generated by one element hencecyclic.

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2 LetG1 = xand G2 = ybe cyclic groups. Assume|G1| = m and |G2| = n. Prove:(m, n) =1 iff every homomorphism : G1 G2is trivial.

Assume m = 1 and n = 1 otherwise every homomorphism is trivial (homomorphismsmust preserve identity som = 1 implies (G1) = ({1G1 }) = {1G2 }and ifn = 1 then

(G1) =G2={1G2 }). Sincem andnare non-negative, we have thatm > 1 andn > 1.Throughout the proof we will freely use | x| = m, |y| = n which is provided by 2.3

Proposition 2. As a consequence of 2.3 Proposition 3, xkn = 1G1 andykn = 1G2 for any

k Z.

Suppose(m, n) =d = 1. Let : G1 G2be a homomorphism.Assume () that is nontrivial, so there is some k Z+ such that (xk) = 1G2 .

SinceG2 = y, there is somel Z+ such that(xk) =y l. Noticen l, otherwise

n| l = p Z+ :y l =ynp = (yn)p =1G2p =1G2

contradictingyl =(xk)=1G2 .Consider (xmk). Notice

(xmk) = ((xm)k) = (1G1k) =(1G1 ) =1G2

since homomorphisms preserve identity. Also, by properties of exponents and homomor-phisms

(xmk) = (xkm ) =((xk)m

) = (xk)m

= (yl)m

=y lm = (ym)l

Together, this implies

1G2=

(

x

mk) = (y

m)l

By 2.3 Proposition 6 (2), since(m, n) =1 we haveG2=ym and 2.3 Proposition 2 implies

|G2|= n = |ym|. By 2.3 Proposition 3,(ym)l =1G2 implies that

|ym| |l n| l

This contradicts our earlier result thatn l (). Therefore, the assumption was incor-rect and it follows that is trivial. Therefore,(m, n) = 1 = every homomorphism: G1 G2is trivial.

We will prove the contrapositive. Assume(m, n) = d = 1. Since both m > 1 and

n > 1, it follows thatd > 1. By definition of gcd and sincem, n, d > 1, there ares, t Z+such that

m= sd and n= td

Define the function: G1 G2

xk ykt

We will show is a non-trivial and well-defined homomorphism.

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Non-trivial: Consider(x) = (x1) = y(1)t = yt. Since d > 1, n > 1, and n = td, itfollows that 0 < t < n. Then by definition of order, |y| = nand 0 < t < nimplies that(x) =yt =1G2so is non-trivial.Well-defined: Letxk, xl G1 fork, l Z

+ be such thatxk = xl. Then by left multipli-cationx kl = 1G1 and 2.3 Proposition 3 impliesn | k l. This means that there is some

p Z such thatk= l + pm. Notice then

(xk) =ykt Def of

=y(l+pm)t k= l+ pm

=y lt+pmt Distributivity in rings:k, l, t Z

=y ltypmt Property of exponents

=y ltyp(sd)t m= sd

=y ltyps(dt) Associativity

=ylt

ypsn

n= td= dt=y lt (yn)ps Property of exponents

=y lt 1G2ps |y|= n

=y lt Identity

=(xl) Def of

Thusxk = x l = (xk) =(xl)so is well-defined.Homomorphism: Let xk, xl G1 for k, l Z

+. Notice by property of exponents anddefinition of

(xkxl) = (xk+l) =y(k+l)t =ykt+lt =yktylt =(xk)(xl)

so is a homomorphism. Since we have shown the contrapositive(m, n)=1 = non-trivial homomorphism : G1 G2, it follows that if every homomorphism : G1 G2is trivial then(m, n) =1.

Altogether, this shows(m, n) =1 every homomorphisms : G1 G2is trivial.

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3

(a) Write the lattice of subgroups ofZ/36Z.

(b) Write the lattice of subgroups ofZ/30Z.

(a)

Figure 1: Subgroup lattice ofZ/36Z

(b)

Figure 2: Subgroup lattice ofZ/30Z

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4 (2.3.11) Find all cyclice subgroups ofD8. Find a proper subgroup ofD8which is notcyclic.

Consider D8=r, s|r4 =s2 =1, rs = sr1. The cyclic subgroups ofD8are:

1 {1}r= r3 {1, r, r2, r3}

r2 {1, r2}s {1, s}rs {1, rs}r2s {1, r2s}r3s {1, r3s}

An example of a proper subgroup which is not cyclic is (from Exercise 2.1.3(a)) S ={1, r2, s, sr2} G. Notice thatS is not isomorphic to any of the subgroup of D8 enu-merated above. We will showSis indeed a valid subgroup.Identity: 1 S by definition.Associativity: Inherited from associative group operation on G.Inverse: Forr2, notice

r4 =r2+2 =r2r2 =1 = (r2)1

S

Fors, clearly

s2 =1 = s1 =s S

Forsr2, using(r2)1

=r2,s1 =s, and the properties of inverses

(sr2)1

= (r2)1

s1 =r2s

From the relationrns= sr n onD8(1.2 Example (6)) and(r2)

1=r2, we have

(sr2)1

=r2s= s(r2)1

=sr2 S

SoScontains the inverse of all its elements.Closure: We construct a multiplication table

1 r2 s sr2

1 1 r2 s sr2

r2 r2 1 sr2 ss s sr2 1 r2

sr2 sr2 s r2 1

Notice everything is contained in S, henceS is closed under the binary operation. SinceS G satisfies the group axioms, S < Gis a non-cyclic proper subgroup.

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Dummit & Foote chapter 2 selected solutions

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