Math 375 Week 6

Math 375

Week 6

6.1 Homomorphisms and Isomorphisms

DEFINITION 1 Let G and G groups and let : G ! G b e a function. Then is a

1 2 1 2

group homomorphism if for every a; b 2 G wehave

ab= ab:

REMARK 1 Notice that the op eration on the left is o ccurring in G while the op er-

ation on the right is o ccurring in G .

REMARK 2 Notice the similarity to the de nition of a linear transformation from

Math 204. I encourage you to lo ok this up in your 204 text. This means

that you can multiply b efore or after you apply the mapping and you

will still get the same answer. This is great, you can't make a mistake

here b ecause the order of op erations mapping versus multiplication

do es not matter.

EXAMPLE 1 Consider the following maps

a Is the mapping : GLn; R ! R by A = det A a homomor-

phism? Yes.

b Let a b e a xed elementofG. Is : G ! G by g = ag a a

homomorphism? [Homework]

c Is the mapping f : R ! R by f x= e a group homomorphism?

Be careful: What are the group op erations in each case?

d Let a b e a xed elementofG. Is : G ! G by g = ag a

homomorphism? No.

e Here's a silly example: Let G and G groups and let : G ! G

1 2 1 2

by g = e for all g 2 G.Obviously, ab= e = e e = bb,

2 2 2 2

so this is a group homomorphism.

f Recall that S is the set of all p ermutations of the set By lab elling

the vertices of an equilateral triangle 1, 2, and 3 as usual, we can

interpret the elements of D as maps from S to S . Match up the

elements of D with their corresp onding elements in S .

3 3 1

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EXAMPLE 2 Extra Credit: Lab el the vertices of a tetrahedron with 1, 2, 3, 4. Let

G b e the set of rotations and re ections of the tetrahedron. Write out

each such rotation as an elementofS .Doyou get all elements of S

4 4

this way? See page 101 and 102 in your text. It turns out that this

pairing is a group homomorphism, indeed, an isomorphism.

THEOREM 2 Basic Prop erties of Homomorphisms If : G ! G is a group

1 2

homomorphism, then

a e = e ;

1 2

1 1

b a = [a] ;

n n

c a = [a] for all n 2 Z;

d if jaj = n, then jaj n, i.e., jaj jaj.

PROOF A Note that e =e e = fe e so that by cancellation e =

1 1 1 1 1 2

e . [Rememb er, this is all taking place in G .]

1 2

PROOF B We prove that something is an inverse by showing that it acts likean

inverse. So

1 1

e = e = aa = aa :

2 1

1 1 1

So f a acts as the inverse to f a, i.e., a = [a] .

PROOF C Homework.

PROOF D Because jaj = n, then a = e .So

n n

e = e = a =[a] :

2 1

By the Corollary on page 73, jaj n.

EXAMPLE 3 Supp ose that : Z ! D is a homomorphism. Can 1 = r ?

3 4 90

SOLUTION No. Because the last part of the theorem we need j1j j1j or but

jr j =46j3= j1j.

EXAMPLE 4 Let's continue with : Z ! D . What can you say ab out this homo-

3 4

morphism?

SOLUTION Since the orders of elements in D are either 1, 2, or 4, the only such

order which divides j1j = 3 is 1. So, 1 = r . But then part b of the

thero em ab ove, 2 = 1 = [1] = r = r . Of course, by part

a of the theorem, 0 = r .Soevery elementinZ must b e mapp ed

0 3

to r . This is the silly example discussed ab ove.

Let's prove something a bit more interesting ab out homomorphisms

of nite cyclic groups. Supp ose that G =is cyclic and : G ! H

6.1 Homomorphisms and Isomorphisms 3

is a group homomorphism. Notice that is completely determined by

where maps a. Because any element g 2 G is of the form g = a ,so

k k

g = a = [a] . Once we know what a is, we know what

is. What are the choices for a?

EXAMPLE 5 Her's the sort of thing I mean. Let's assume : Z ! Z is a homomor-

8 4

phism. Supp ose that 1 = 3 . Then the rest of is now completely

8 4

determined b ecause Z =< 1 >.We b ecause is a homomorphism

Or one could use j = j 1 ! j 3 to get the same answers.

8 8 4

Ok, let's generalize nite case rst.

LEMMA 3 Let G =b e a cyclic group of order n. Let : G ! H be a

i i

function such that a =a for all i.Ifjaj n, then is a group

homomorphism.

PROOF Note from parts c and d of the previous theorem, if is a homomor-

phism, then these two prop erties must b e true. This says that these two

prop erties suce to make a homomorphism when G is cyclic.

Let jaj = m. Then we are given that m j n,son = md for some

d 2 Z. Let x; y 2 G.Wemust show that xy = xy . But G

j k j k

is cyclic so x = a and y = a with 0 k;j

j +k mo d n

a .So

j +k mo d n j +k mo d nmodm

xy = a =[a] :

The mo d m is necessary since jaj = m. On the other hand,

j k j mo d m k mo d m j +k mod m

xy = a a =[a] [a] =[a] :

So it all b oils down to whether j + k mo d nmodm =j + k mod m.

By the division algorithm, wemay write

j + k = qn + s 0 s< n:

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So j + k mo d nmodm = s mo d m Since n = dm,wehave j + k =

qn + s = q dm+ s =qdm + s: But then j + k modm = s mo d m,

to o. So

j +k mo d nmodm s mo d m

xy = [a] =[a]

j +k mod m

=[a]

j mo d m k mo d m

=[a] [a]

= xy :

When G is an in nite cyclic group the pro of is even easier.

LEMMA 4 Let G = b e an in nite cyclic group. Let : G ! H b e a function

i i

such that a =a for all i. Then is a group homomorphism.

Again, let x; y 2 G.Wemust show that xy = xy . But G is

j k

cyclic so x = a and y = a There are two cases. If jaj = 1, then by

assumption

j +k j +k j k

xy = a = [a] =[a] [a] = xy :

If jaj = n, then by assumption

Another crucial fact is that comp osites of homomorphisms are ho-

momorphisms.

LEMMA 5 Let : G ! H , and : H ! K b e group homomorphisms. Then so is

the comp osite, : G ! K .

PROOF Let a; b 2 G. Then since b oth maps are homomorphisms,

ab= ab = ab

= a f b= [ a][ b]:

DEFINITION 6 If in addition a homomorphism : G ! G is b oth injective and

1 2

surjective then is called a group isomorphism. The two groups are

said to b e isomorphic and this is denoted by G G .

1 2

REMARK Note that to provetwo groups are isomorphic, wemust 1 nd a map-

ping : G ! G ; 2 show that is injective; 3 show that is

1 2

surjective; and 4 show that is a homomorphism.

6.1 Homomorphisms and Isomorphisms 5

EXAMPLE a For homework, if G is a group and a is a xed elelmentofG, then

the mapping : G ! G by g = ag a is an injective, surjective

homomorphism. Thus is an isomorphism.

b Wesaw that if G were a group and a was a xed elelmentofG,

then the mapping : G ! G by g = ag was an injective and

surjective. Let's check to see if it is an isomorhism. gh = ag h,

while gh= ag ah. The two are not equal, and thus is

not an isomorphism.

c The simplest example is the identity mapping. Let G b e a group

and let i : G ! G by i g = g .We know that i is injective and

G G G

surjective and clearly

i ab= ab = i ai b:

G G G

So i is an isomorphism and G G. In other words, is a re exive

= =

relation. Is it an equivalence relation?

d Another imp ortant mapping for abelian groups is : G ! G by

g = g . This map is injective: Let a; b 2 G. Then

1 1

a= b a = b a = b:

is surjective: Let c 2 G co domain. Find a 2 G so that a= c.

But

1 1

a= c a = c a = c :

So why is ab elian necessary here? When wechek the homomorphism

prop erty,

1 1 1 1 1

ab= ab = b a = a b = ab:

But ths is only p ossible b ecuase the group is ab elian.

G then jG j = jG j. LEMMA 7 If G

2 1 2 1

PROOF There's a one-to-one onto map b etween the two sets. So counting the

elements of one set simultaneously conts the elements of the other.

EXAMPLE 7 Can Q b e isomorphic to Z ? There is another reason that there is no

8 10

isomorphism b etween these two groups.

THEOREM 8 Let G = b e a cyclic group.

a If jGj = n, then Z G.

b If jGj = 1, then Z G. =

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PROOF In the rst case, note that G =. De ne de ne : Z ! G by

k =a for any k 2 Z . The map is injective since

k j

k = j a = a n j j k

whichby Theorem 4.1. But this means that j = k mo d n. The map is

surjective, obviously. Finally, it is a homomorphism by the lemma we

proved earlier, since n j n.

In the second case, de ne : Z ! G by k = a . The map is

injective since

k j

j = k a = a k = j

by Theorem 4.1 since jaj = 1. The map is surjective, obviously. Finally,

it is a homomorphism si if j; k 2 Z, then

j +k j k

j + k = a = a a =jk:

EXAMPLE 8 fi; 1; i; 1g Z since b oth are cyclic of order four. What would the

isomorphism apping b e here?

U 7 since b oth are cyclic of order 6. Use 1 ! 3. EXAMPLE 9 Z

THEOREM 9 Prop erties of Group Isomorphisms Let : G ! G b e a group

1 2

isomorphism. Then in addition to the prop erties of the previous theorem:

a : G ! G is an isomorphism;

2 1

b jaj = jaj;

c G is cyclic if and only if G is cyclic;

1 2

d a; b 2 G commute if and only if a;b 2 G commute;

1 2

e G is ab elian if and only if G is ab elian.

1 2

f If H G , then H = fh j h 2 H g is a subgroup of G .

1 2

A Since is an isomorphism, it is surjective and injective, so : G ! G

2 1

exists and is injective and surjective. We only need to show that it is

a group homomorphism. So take g ;h 2 G . Wemust show that

2 2 2

1 1 1 1 1

g h = g h . Let g = g and h = h . Then

2 2 2 2 2 1 2 1

g =g and h = h . Since is a homomorphism, g h =

1 2 1 2 1 1

g h =g h . Therefore,

1 1 2 2

1 1 1

g h =g h = g h :

2 2 1 1 2 2

B Both and are homomorphisms. So we know that jaj jaj and

jaj. Therefore, the orders are equal. What happ ens j a = jaj

if jaj = 1?

6.1 Homomorphisms and Isomorphisms 7

C Supp ose G =is cyclic. Then let a=b 2 G .We will show

1 2

that G =. Let g 2 G . Because surjective, there is an element

2 2 2

g 2 G so that g = g . But G =,sog = a for some k 2 Z.

1 1 1 2 1 1

Therefore,

k k k

g = g =a = [a] = b :

2 1

Therefore G is cyclic. If G is cyclic, then so is G . Simply use the fact

2 2 1

that is an isomorphism.

D This is a homework problem.

E Follows from g and the fact that is onto.

F Use the one-step test. Let x; y 2 H . Wemust show that xy 2 H .

But there are elements in a; b 2 H so that a=x and b= y .

Moreover, since H is a subgroup, then ab 2 H .So

1 1 1 1

xy = a[b] = ab =ab 2 H ;

since ab 2 H .

EXAMPLE a Z is not isomorphic D ,even though b oth have the smae number

6 3

of elements. [Give three di erent reasons!]

b Z is not isomorphic to R since one is cyclic and the other is not.

One has an element of order 2, the other do es not.

c U 12 is not isomorphic to Z even though b oth are ab elian and

have the same numb er of elements. U 12 = f1; 5; 7; 11g is not

cyclic. However, Z = fi; i; 1; 1g since b oth are cyclic.

Z is not isomorphic to V since the latter is not cyclic. What

4 4

ab out the group of motions of a rectangle? Of a rhombus? Is either

isomorphic to Z . Explain.

d C is not isomorphic to R .If were such an isomorphism, and

i= x, then jxj = jij = jij = 4. But the only elements of nite

order in R are 1 and 1 whichhave order 1 and 2, resp ectively.

e R is not isomorphic to R, b ecause 1inR has order 2 and no

elementinR has order 2. Recall, however, that we know that R

is isomorphic to R; use =lnx.

f Show that D is not isomorphic to S . Both ahve order 24 and are

12 4

not ab elian. But the former has an element of order 12, while the

later has no elements whose order is greater than 4.

g Is D isomorphic to Q ? Find out by lo oking at their tables. No.

4 8

Check the numb er of elements of order 4 in each group.

THEOREM 10 Cayley Every group is isomorphic to a group of p ermutations.

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See text. The result is interesting but very hard to use in practice, so

we will ignore it temp orarily.

DEFINITION 11 An isomorphism : G ! G from a group to itself is called an automor-

phism.

EXAMPLE 11 Wehave seen that for ab elian groups the mapping : G ! G by g =

g is an isomorphism, hence it is an automorphism. Similarly, for

any group G and any element a 2 G, the mapping : G ! G by

 g = a ga is an isomorphism. Hence is an automorphism. is

a a a

called the inner automorphism of G induced by a.

Let's lo ok at a sp ec c example of an inner automorphism.

EXAMPLE 12 Let =1; 2; 3 2 S . What is the mapping : S ! S ? Well,

3 3 3

=3; 2; 1, so

x ! x

e = 1 ! 3; 2; 111; 2; 3 = 1

1; 2 ! 3; 2; 11; 21; 2; 3 = 2; 3

1; 3 ! 3; 2; 11; 31; 2; 3 = 1; 2

2; 3 ! 3; 2; 12; 31; 2; 3 = 1; 3

1; 2; 3 ! 3; 2; 11; 2; 31; 2; 3 = 1; 2; 3

3; 2; 1 ! 3; 2; 13; 2; 11; 2; 3 = 3; 2; 1

DEFINITION 12 The set of all automorphisms of G is called AutG and the set of all

inner automorphisms is called Inn G.

THEOREM 13 Let G b e a group. Then AutG and Inn G are also groups. They are

b oth subgroups of S the set of p ermutations of elements of G.

PROOF We'll show that AutG is a subgroup of S . Inn G is less imp ortant

at this p oint. AutG is simply the set of injecitve, surjective maps from

G to itself that are also group homomorphisms. Let ; 2 AutG:

1 1

Show that 2 AutG. All we need to do is show that is a

homomorphism. But since is an automorphism, it is an isomorphism

1 1

so is an isomorphsim why. So b oth and are homomorphisms

from G to G, hence so is there comp osite.