Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

APP. B] ALGEBRAIC SYSTEMS 457

B.18.SupposeF:G→G′is a group homomorphism. Prove: (a)f(e)=e′;(b)(fa−^6 )=f(a)−^1.

(a) Sincee=eeandfis a homomorphism, we have

f(e)=f (ee)=f(e)f(e)

Multiplying both sides byf(e)−^1 gives us our result.

(b) Using part (a) and thataa−^1 =a−^1 a=e, we have

e′=f(e)=f(aa−^1 )=f(a)f(a−^1 ) and e′=f(e)=f(a−^1 a)=f(a−^1 )f (a)

Hencef(a−^1 )is the inverse off(a); that is,f(a−^1 )=f(a)−^1.

B.19.Prove Theorem B.9: Letf:G→G′be a homomorphism with kernelK. ThenKis a normal subgroup

ofG, andG/Kis isomorphic to the image off. (Compare with Problem B.5, the analogous theorem for

semigroups.)

Proof that K is normal: By Problem B.18,f(e)=e′,soe∈K. Now supposea,b∈Kandg∈G. Then

f(a)=e′andf(b)=e′. Hence

f(ab)=f(a)f(b)=e′e′=e′

f(a−^1 )=f(a)−^1 =e′−^1 =e′

f(gag−^1 )=f(g)f(a)f(g−^1 )=f(g)e′f(g)−^1 =e′

Henceab,a−^1 , andgag−^1 belong toK,soKis a normal subgroup.

Proof thatG/K∼=H, where H is the image of f: Letφ:G/K→Hbe defined by

φ(Ka)=f(a)

We show thatφis well-defined, i.e., ifKa=Kbthenφ(Ka)=φ(Kb). SupposeKa=Kb. Thenab−^1 ∈K

(Problem B.57). Thenf(ab−^1 )=e′, and so

f(a)f(b)−^1 =f(a)f(b−^1 )=f(ab−^1 )=e′

Hencef(a)=f(b), and soφ(Ka)=φ(Kb). Thusφis well-defined.

We next show thatφis a homomorphism:

φ(KaKb)=φ(Kab)=f(ab)=f(a)f(b)=φ(Ka)φ(Kb)

Thusφis a homomorphism. We next show thatφis one-to-one. Supposeφ(Ka)=φ(Kb). Then

f(a)=f(b) or f(a)f(b)−^1 =e′ or f(a)f(b−^1 )=e′ or f(ab−^1 )=e′

Thusab−^1 ∈K, and by Problem B.57 we haveKa=Kb. Thusφis one-to-one. We next show thatφis onto. Let

h∈H. SinceHis the image off, there existsa∈Gsuch thatf(a)=h. Thusφ(Ka)=f(a)=h, and soφis

onto. ConsequentlyG/K∼=Hand the theorem is proved.

RINGS, INTEGRAL DOMAINS, FIELDS

B.20.Consider the ring Z 10 ={ 0 , 1 , 2 ,..., 9 }of integers modulo 10. (a) Find the units of Z 10.

(b) Find−3,−8, and 3−^1 .(c) Letf(x)= 2 x^2 + 4 x+4. Find the roots off(x)overZ 10.

(a) By Problem B.78 those integers relatively prime to the modulusm=10 are the units inZ 10. Hence the units are

1, 3, 7, and 9.

(b) Recall that−ain a ringRis the element such thata+(−a)=(−a)+a=0. Hence− 3 =7 since 3+ 7 = 7 + 3 = 0

inZ 10. Similarly− 8 =2. Recall thata−^1 in a ringRis the element such thata·a−^1 =a−^1 ·a=1. Hence

3 −^1 =7 since 3· 7 = 7 · 3 =1inZ 10.