Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

MAT301H1S Lec5101 Burbulla

Week 5 Lecture Notes

Winter 2020

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

What is a Homomorphism?

Definition: Let G and H be groups and suppose f : G −→ H is a

function such that for all x, y ∈ G ,

f (xy ) = f (x)f (y ).

Then f is called a group homomorphism, or just homomorphism

for short. In words, we say “f preserves products.”

Note: to calculate xy we are using the operation in G , but to

calculate f (x)f (y ) we are using the operation in H.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Examples

1. f : R∗ −→ R∗ by f (x) = |x|
f (xy ) = |xy | = |x||y | = f (x)f (y ).
2. f : R∗ −→ R∗ by f (x) = x 2
f (xy ) = (xy )2 = x 2 y 2 = f (x)f (y ).
3. f : Z −→ Zn by f (m) = m mod n, with addition. Use the
division algorithm to write

m1 = p1 n + q1 , m2 = p2 n + q2 .

Then m1 + m2 = (p1 + p2 )n + q1 + q2 , so
f (m1 +m2 ) = (q1 +q2 ) mod n ≡ q1 mod n+q2 mod n = f (m1 )+f (m2 )

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Not All Functions are Homomorphisms

Let f : R −→ R by f (x) = x 2 . The operation is addition. Then

f (x + y ) = (x + y )2 = x 2 + 2xy + y 2

but
f (x) + f (y ) = x 2 + y 2 .
So it is not true that f (x + y ) = f (x) + f (y ) for all x, y ∈ R, so f
is not a homomorphism.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Some More Interesting Examples

The determinant function, det : GL(n, R) −→ R∗ , is a
homomorphism, since

det(AB) = det(A) det(B).

Define sgn : Sn −→ {1, −1} by


1 if σ is an even permutaton
sgn (σ) =
−1 if σ is an odd permutaton

You have to check that for all α, β ∈ Sn ,

sgn (αβ) = sgn (α) sgn (β),

which boils down to checking four cases, namely even plus even is
even, even plus odd is odd, odd plus even is odd, and odd plus odd
is even.
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla
 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Properties of Homomorphisms: Th 10.1 First Three Parts

Let f : G −→ H be a homomorphism. Then

1. f (eG ) = eH
Proof:
x = xeG ⇒ f (x) = f (xeG ) = f (x)f (eG ) ⇒ eH = f (eG ).
2. f (x n ) = (f (x))n , for all x ∈ G and all n ∈ Z
Proof: f (x 2 ) = f (x x) = f (x) f (x) = (f (x))2 , etc for positive
n. If n = −1, then
xx −1 = eG ⇒ f (xx −1 ) = f (eG ) ⇒ f (x)f (x −1 ) = eH ⇒
f (x −1 ) = (f (x))−1 .
3. If |x| is finite, then |f (x)| divides |x|
Proof: let |x| = n. Then
x n = eG ⇒ f (x n ) = f (eG ) ⇒ (f (x))n = eH ⇒
|f (x)| divides n.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Kernels and Images

If f : G −→ H is a homomorphism, define
1. ker(f ) = {x ∈ G | f (x) = eH }
2. im (f ) = {f (x) | x ∈ G } = f (G )
Then
ker(f ) ≤ G and im (f ) ≤ H.
Proof: (for kernel)
I f (eG ) = eH ⇒ eG ∈ ker(f ).
I x, y ∈ ker(f )
⇒ f (xy −1 ) = f (x)f (y −1 ) = f (x) (f (y ))−1 = eH eH = eH ,
so xy −1 ∈ ker(f ).
I Proof that im (f ) ≤ H is left as an exercise.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Examples

1. For det : GL(n, R) −→ R∗ ,

ker(det) = {A ∈ GL(n, R) | det(A) = 1} = SL(n, R).
2. For sgn : Sn −→ {1, −1}
ker(sgn ) = {σ ∈ Sn | sgn (σ) = 1} = An .
3. For f : Z −→ Zn by f (m) = m mod n, with addition,

ker(f ) = {m | m ≡ 0 mod n} = < n >

and
im (f ) = {m mod n | m ∈ Z} = Zn .

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Proper Definition of Determinant

Let A = (aij ) be an n × n matrix. Here is the “proper” definition
of det(A) :
X
det(A) = sgn (σ) a1σ(1) a2σ(2) · · · · · anσ(n) .
σ∈Sn

For example, if n = 3, the even permutations are (1), (123), (132)

and the odd permutations are (12), (23), (31). Then
 
a11 a12 a13
det  a21 a22 a23  =
a31 a32 a33

a11 a22 a33 +a12 a23 a31 +a13 a21 a32 −a12 a21 a33 −a11 a23 a32 −a13 a22 a31 .

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Properties of Subgroups Under Homomorphisms

First three parts of Theorem 10.2: let f : G −→ H be a
homomorphism and let K be a subgroup of G . Then
1. f (K ) = {f (k) | k ∈ K } is a subgroup of H.
2. If K is cyclic then f (K ) is cyclic.
3. If K is Abelian then f (K ) is Abelian.
Proof: 1. left as an exercise.
2. Suppose K = hki. Then x ∈ K ⇒ x = k n . Then

f (x) = f (k n ) = (f (k))n ,

which means that f (K ) = hf (k)i.

3. Suppose xy = yx for all x, y ∈ K . Then

f (xy ) = f (yx) ⇔ f (x)f (y ) = f (y )f (x),

which means all elements in f (K ) commute.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Example
 
cos θ − sin θ
Let f : R −→ GL(2, R) by f (θ) = . f is a
sin θ cos θ
homomorphism because f (α)f (β) = f (α + β) :
    
cos α − sin α cos β − sin β cos(α + β) − sin(α + β)
= ,
sin α cos α sin β cos β sin(α + β) cos(α + β

using appropriate trig identities. Check that ker(f ) = h2πi and

im (f ) is the group of 2 × 2 rotation matrices. In particular, R is
Abelian, so im (f ) is Abelian and im (f ) is an Abelian subgroup of
GL(2, R). If the positive integer n is fixed and
   
2π 2πk
K= = |k ∈Z ,
n n
then f (K ) = hf (2π/n)i ≤ Dn , consisting of all rotations of a
regular n-gon.
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla
 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

One-to-one and Onto Homomorphisms

Let f : G −→ H be a group homomorphism.
Definition: f is called one-to-one if

f (x1 ) = f (x2 ) ⇒ x1 = x2 .

Defintion: f is called onto if every element in H is in im (f ).

Theorem: let f : G −→ H be a group homomorphism.
I f is one-to-one if and only if ker(f ) = {eG }
I f is onto if and only if im (f ) = H

Proof: (for one-to-one) suppose f is one-to-one: Let x ∈ ker(f ).

Then f (x) = eH and f (eG ) = eH . Therefore x = eG .
Now suppose ker(f ) = {eG }, and let f (x1 ) = f (x2 ). Then

f (x1 x2−1 ) = f (x1 )f (x2−1 ) = f (x1 )(f (x2 ))−1 = f (x1 )(f (x1 ))−1 = eH .

Thus x1 x2−1 ∈ ker(f ) ⇒ x1 x2−1 = eG ⇒ x1 = x2 .

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

What is an Isomorphism?

Let f : G −→ H be a group homomorphism.

Definition: f is called an isomorphism if f is one-to-one and onto.
Thus f is an isomorphism if ker(f ) = {eG } and im (f ) = H.
Note: every isomorphism f : G −→ H has an inverse
f −1 : H −→ G , defined by

f −1 (x) = y ⇔ x = f (y ),

which is also an isomorphism.

Definition: if there is an isomorphism f : G −→ H, then we say
the groups G and H are isomorphic, and we write

G ≈ H.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Example 1
  
1 m
Let H = ∈ GL(2, R) | m ∈ Z and define f : Z −→ H
0 1
by  
1 m
f (m) = .
0 1
Then f is an isomorphism:
1. f is a homomorphism since
    
1 m+n 1 m 1 n
f (m+n) = = = f (m)f (n)
0 1 0 1 0 1

 
1 0
2. f is one-to-one since f (m) = ⇒m=0
0 1
3. f is onto since obviously im (f ) = H.
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Example 2

Let G = hai be a cyclic group of order n. Then G ≈ Zn .

Proof: for k ∈ Z, define f : G −→ Zn by f (ak ) = k mod n.
I f is well-defined: aj = ak ⇒ n divides j − k ⇒ j ≡ k mod n.
I f is a homomorphism: f (aj ak ) = f (aj+k ) = (j + k) mod n;
and f (aj ) + f (ak ) = j mod n + k mod n = (j + k) mod n.
I f is one-to-one:
f (ak ) = 0 ⇒ k ≡ 0 mod n ⇒ k = qn ⇒ ak = (an )q = eG .
I f is onto: this is obvious since k ∈ Z, so im (f ) = Zn .
Alternate Proof: use g : Zn −→ G defined by g (m) = am .
(g = f −1 .) This makes the algebra easier since g is obviously
well-defined.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Properties of Isomorphisms
Ref: Theorems 6.2 and 6.3. Let f : G −→ H be an isomorphism.
Then
1. f −1 : H −→ G is also an isomorphism
2. f (eG ) = eH
3. for all x ∈ G , f (x n ) = (f (x))n
4. x, y commute in G if and only if f (x), f (y ) commute in H
5. G = hxi if and only if H = hf (x)i
6. for all x ∈ G , |x| = |f (x)|
7. if G and H are finite groups, then |G | = |H|
8. G is Abelian if and only if H is Abelian
9. G is cyclic if and only if H is cyclic
10. f (Z (G )) = Z (H)

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Some Proofs:
Properties 2 and 3 are true for all homomorphisms.
4. xy = yx ⇒ f (xy ) = f (yx) ⇒ f (x)f (y ) = f (y )f (x); this is true
for any homomorphism. But if f is an isomorphism, then
f (x)f (y ) = f (y )f (x) ⇒ f (xy ) = f (yx) ⇒ xy = yx, since f is
one-to-one
6. let |x| = n, |f (x)| = m. For any homomorphism f we know m
divides n. If f is also an isomorphism: (f (x))m = eH ⇒
f (x m ) = eH ⇒ x m = eG , since f is one-to-one. Thus n divides m.
So m = n.
8. follows from 4
9. follows from 5
10. x ∈ Z (G ) ⇒ xg = gx, for all g ∈ G ⇒ f (x)f (g ) = f (g )f (x)
⇒ f (x) ∈ Z (im (f )). So, for any homomorphism,
f (Z (G )) ⊂ Z (im (f )). If f is an isomorphism, then im (f ) = H. For
the other inclusion, repeat the argument for f −1 : H −→ G .
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla
 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Example 3
Show U(8) ≈ {(1), (12)(34), (13)(24), (23)(14)}
Solution: U(8) = {1, 3, 5, 7}. Define
f : U(8) −→ {(1), (12)(34), (13)(24), (23)(14)} by

f (1) = (1), f (3) = (12)(34), f (5) = (13)(24), f (7) = (23)(14).

Now consider the multiplication tables of both groups:

· 1 3 5 7 ◦ (1) (12)(34) (13)(24) (23)(14)
1 1 3 5 7 (1) (1) (12)(34) (13)(24) (23)(14)
3 3 1 7 5 (12)(34) (12)(34) (1) (23)(14) (13)(24)
5 5 7 1 3 (13)(24) (13)(24) (23)(14) (1) (12)(34)
7 7 5 3 1 (23)(14) (23)(14) (13)(24) (12)(34) (1)
Observe that f maps the multiplication table on the left precisely
to the multiplication table on the right: i.e. f (xy ) = f (x)f (y ). We
say, isomorphic groups have the same group structure.
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Example 4

Finite groups that are isomorphic have the same order. But groups
can have the same order and not be isomorphic. For example, here
are three groups of order 12: Z12 , D6 , and A4 , no two of which are
isomorphic. There are many ways to show this:
I the largest order of any element in Z12 , D6 or A4 is 12, 6, or
3, respectively.
I compare the number of elements of order 2: Z12 has 1, D6
has 7, and A4 has 3.
I Z12 is Abelian and cyclic; the biggest cyclic subgroup D6 has
is of order 6; the biggest cyclic subgroup A4 has is of order 3.
There are only five non-isomorphic groups of order 12: the three
above, plus one other Abelian group and one other non-Abelian
group.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Cayley’s Theorem
Theorem 6.1: Every group is isomorphic to a group of
permutations. If |G | = n, then G is isomorphic to a subgroup of
Sn .
Proof: given a group G we shall construct a group of
permutations, H, and then show that G ≈ H. For g ∈ G , define
Tg : G −→ G by Tg (x) = gx. Since Tg (g −1 y ) = y , Tg is onto;
since Tg (x) = Tg (y ) ⇒ gx = gy ⇒ x = y , Tg is one-to-one.
Thus Tg is a permutation of all the elements of G . Let

H = {Tg | g ∈ G },

with function composition, and define φ : G −→ H by

φ(g ) = Tg .

We claim φ is an isomorphism:
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

1. φ is a homomorphism: φ(ab) = φ(a) ◦ φ(b). To show two

functions are equal you have to evaluate them at a point.
(φ(ab))(x) = Tab (x) = abx and
(φ(a) ◦ φ(b))(x) = Ta (Tb (x)) = Ta (bx) = abx.
2. φ is one-to-one: φ(a) = φ(b) ⇒ Ta (x) = Tb (x), for all x ∈ G
⇒ ax = bx ⇒ a = b.
3. φ is onto: by definition of φ. For any Tg ∈ H, φ(g ) = Tg .
Finally, if |G | = n, then you can number the elements of G as
x1 , x2 , . . . , xn and each element Tg ∈ H can be considered as a
permutation σ ∈ Sn defined by

Tg (xi ) = xσ(i) .

Note: Example 3 above showed that U(8) is isomorphic to a

subgroup of S4 .
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla
 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

What Is An Automorphism?

Defintion: if f : G −→ G is an isomorphism, of G to itself, then

we call f an automorphism of G .

Every group G has at least one automorphism, the identity map

i : G −→ G , defined by i(g ) = g .

The map f : G −→ G defined by f (x) = x −1 is an automorphism

only if G is Abelian:

f (xy ) = f (x)f (y ) ⇔ (xy )−1 = x −1 y −1 ⇔ y −1 x −1 = x −1 y −1 .

This last statement will be true if G is Abelian.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Example 5
Define f : GL(n, R) −→ GL(n, R) by f (A) = (A−1 )T . Then f is an
automorphism of GL(n, R).
1. f is a homomorphism:
f (AB) = ((AB)−1 )T = (B −1 A−1 )T = (A−1 )T (B −1 )T = f (A)f (B).

2. f is one-to-one:
f (A) = I ⇒ (A−1 )T = I ⇒ A−1 = I T = I ⇒ A = I −1 = I

3. f is onto: observe that

   −1 T
f (AT )−1 = (AT )−1 = (AT )T = A

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

What Is an Inner Automorphism?

Theorem: for g ∈ G , φg : G −→ G , defined by

φg (x) = gxg −1

is an automorphism of G . Proof:
1. φg is a homomorphism:

φg (x)φg (y ) = (gxg −1 )(gyg −1 ) = gxyg −1 = φg (xy )

2. φ is one-to-one: φg (x) = e ⇒ gxg −1 = e ⇒ x = g −1 g = e.

3. φg is onto: for any x ∈ G , φg (g −1 xg ) = g (g −1 xg )g −1 = x
Definition: φg is called an inner automorphism of G , or to be
more specific, the inner automorphism of G induced by g .
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Example 6
Note 1: if G is Abelian, then φg is always the identity function,
because φg (x) = gxg −1 = gg −1 x = x.
Note 2: if g ∈ Z (G ), then φg is the identity function, because
φg (x) = gxg −1 = xgg −1 = x.
Thus the only interesting inner automorphisms of a group are φg
for g ∈
/ Z (G ).  
2 1
Example: let A = ; then φA : GL(2, R) −→ GL(2, R) is
7 4
       
a b a b 2 1 a b 4 −1
φA =A A−1 =
c d c d 7 4 c d −7 2
 
8a + 4c − 14b − 7d −2a − c + 4b + 2d
=
28a + 16c − 49b − 28d −7a − 4c + 14b + 8d

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

 Chapter 10: Group Homomorphisms
Chapter 6: Isomorphisms

Groups of Automorphisms
Theorem 6.4: let Aut (G ) be the set of all automorphisms of G ;
let Inn (G ) be the set of all inner automorphisms of G . Then both
Aut (G ) and Inn (G ), with operation composition of functions, are
groups, and Inn (G ) ≤ Aut (G ).
Proof: (some) the identity map, i(x) = x, is in Aut (G ), so it is
non-empty. If f , g ∈ Aut (G ), we need to show f ◦ g −1 ∈ Aut (G ) :
1. f ◦ g −1 is a homomorphism: (f ◦ g −1 )(xy ) =
f ((g −1 )(xy )) = f (g −1 (x)g −1 (y )) = f (g −1 (x))f (g −1 (y )) =
(f ◦ g −1 )(x)(f ◦ g −1 )(y ).
2. f ◦ g −1 is one-to-one: f (g −1 (x)) = e ⇒ g −1 (x) = e ⇒ x = e
3. f ◦ g −1 is onto: for any x ∈ G , check that
(f ◦ g −1 )(g (f −1 (x)) = x.
This shows that Aut (G ) contains the identity and is closed under
composition and inverses. Finally, composition is associative: it’s
always true that f ◦ (g ◦ h) = (f ◦ g ) ◦ h. (Similarly for Inn (G ).)
Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Example 7

Aut (Z42 ) ≈ U(42) : let f ∈ Aut (Z42 ). Since Z42 = h1i, f is

completely determined by what it does to 1. Let f (1) = m. Since f
is an isomorphism, |f (1)| = |1| = 42, so there are only twelve
possibilities for f :

f (1) = 1, 5, 11, 13, 17, 19, 23, 25, 29, 31, 37 or 41.

That is, f (1) = m ⇔ m ∈ U(42). Let fm be the autormorphism of

Z42 such that fm (1) = m. Check that the correspondence
φ : U(42) −→ Aut (Z42 ), given by φ(m) = fm , is an isomorphism.

This is a special case of Theorem 6.5: for every positive integer n,

Aut (Zn ) ≈ U(n).

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla