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Chapter 9
Homomorphisms and the Isomorphism
Theorems
9.1 Homomorphisms
Let G (^) 1 and G (^) 2 be groups. Recall that : G (^) 1! G (^) 2 is an isomorphism i↵
(a) is one-to-one,
(b) is onto, and
(c) satisfies the homomorphic property.
We say that G (^) 1 is isomorphic to G (^) 2 and write G (^) 1 G (^) 2 if such a exists. Loosely speaking, two groups are isomorphic if they have the “same structure.” What if we drop the one-to- one and onto requirement?
Definition 9.1. Let ( G (^) 1 , ⇤) and ( G (^) 2 , ) be groups. A function : G (^) 1! G (^) 2 is a homomor- phism i↵ satisfies the homomorphic property:
( x ⇤ y ) = ( x ) ( y )
for all x, y 2 G (^) 1. At the risk of introducing ambiguity, we will usually omit making explicit reference to the binary operations and write the homomorphic property as
( xy ) = ( x ) ( y ).
Group homomorphisms are analogous to linear transformations on vector spaces that one encounters in linear algebra. Figure 9.1 captures a visual representation of the homomorphic property. We encoun- tered this same representation in Figure 5.6. If ( x ) = x^0 , ( y ) = y^0 , and ( z ) = z^0 while z^0 = x^0 y^0 , then the only way G (^) 2 may respect the structure of G (^) 1 is for
( x ⇤ y ) = ( z ) = z^0 = x^0 y^0 = ( x ) ( y ).
⇤ y
x z !
y^0
x^0 z^0
Figure 9.
Exercise 9.2. Define : Z 3! D 3 via ( k ) = r k^. Prove that is a homomorphism and then determine whether is one-to-one or onto. Also, try to draw a picture of the homomor- phism in terms of Cayley diagrams.
Exercise 9.3. Let G and H be groups. Prove that the function : G ⇥ H! G given by ( g, h ) = g is a homomorphism. This function is an example of a projection map.
There is always at least one homomorphism between two groups.
Theorem 9.4. Let G (^) 1 and G (^) 2 be groups. Define : G (^) 1! G (^) 2 via ( g ) = e (^) 2 (where e (^) 2 is the identity of G (^) 2 ). Then is a homomorphism. This function is often referred to as the trivial homomorphism or the 0-map.
Back in Section 5.5, we encountered several theorems about isomorphisms. However, at the end of that section we remarked that some of those theorems did not require that the function be one-to-one and onto. We collect those results here for convenience.
Theorem 9.5. Let G (^) 1 and G (^) 2 be groups and suppose : G (^) 1! G (^) 2 is a homomorphism.
- If e (^) 1 and e (^) 2 are the identity elements of G (^) 1 and G (^) 2 , respectively, then ( e (^) 1 ) = e (^) 2.
- For all g 2 G (^) 1 , we have ( g ^1 ) = [ ( g )] ^1.
- If H G (^) 1 , then ( H ) G (^) 2 , where
( H ) := { y 2 G (^) 2 | there exists h 2 H such that ( h ) = y }.
Note that ( H ) is called the image of H. A special case is when H = G (^) 1. Notice that is onto exactly when ( G (^) 1 ) = G (^) 2.
The next two theorems tell us that under a homomorphism, the order of the image must divide the order of the preimage.
Theorem 9.6. Let G (^) 1 and G (^) 2 be groups and suppose : G (^) 1! G (^) 2 is a homomorphism. If G (^) 1 is finite, then | ( G (^) 1 )| divides | G (^) 1 |.
Theorem 9.7. Let G (^) 1 and G (^) 2 be groups and suppose : G (^) 1! G (^) 2 is a homomorphism. If g 2 G (^) 1 such that | g | is finite, then | ( g )| divides | g |.
Every homomorphisms has an important subset of the domain associated with it.
9.2 The Isomorphism Theorems
We begin with a theorem.
Theorem 9.19. Let G be a group and let H E G. Then the map : G! G/H given by ( g ) = gH is a homomorphism with ker( ) = H. This map is called the canonical projection map.
The upshot of Theorems 9.11 and 9.19 is that kernels of homomorphisms are always normal and every normal subgroup is the kernel of some homomorphism. The next theorem is arguably the crowning achievement of the course.
Theorem 9.20 (The First Isomorphism Theorem). Let G (^) 1 and G (^) 2 be groups and suppose : G (^) 1! G (^) 2 is a homomorphism. Then
G (^) 1 / ker( ) ( G (^) 1 ).
If is onto, then G (^) 1 / ker( ) G (^) 2_._
Exercise 9.21. Let : Q (^) 8! V 4 be the homomorphism described in Exercise 9.14. Use the First Isomorphism Theorem to prove that Q (^) 8 / h 1 i V 4.
Exercise 9.22. Define : S (^) n! Z 2 via
0 , even 1 , odd_._
Use the First Isomorphism Theorem to prove that S (^) n /A (^) n Z 2.
Exercise 9.23. Use the First Isomorphism Theorem to prove that Z / 6 Z Z 6. Attempt to draw a picture of this using Cayley diagrams.
Exercise 9.24. Use the First Isomorphism Theorem to prove that (Z 4 ⇥ Z 2 ) / ({ 0 } ⇥ Z 2 ) Z 4.
We finish the chapter by listing a few of the remaining isomorphism theorems, but we won’t prove these in this course.
Theorem 9.25 (The Second Isomorphism Theorem). Let G be a group with H G and N E G. Then
- HN := { hn | h 2 H, n 2 N } G ;
- H \ N E H ;
- H/H \ N HN /N.
Theorem 9.26 (The Third Isomorphism Theorem). Let G be a group with H, K E G and K H. Then G/H ( G/K ) / ( H/K ).
