Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

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§ 3.7 Homomorphisms

 
Definition 3.7.1. Let G1 and G2 be groups, and let µ : G1 -> G2 be a function. Then µ is said to be a group homomorphism if

µ(ab) = µ(a) µ(b)

for all a,b in G1.

 
Example 3.7.1. (Exponential functions for groups) Let G be any group, and let a be any element of G. Define µ : Z -> G by µ(n) = an, for all n in Z. This is a group homomorphism from Z to G.
If G is abelian, with its operation denoted additively, then we define µ : Z -> G by µ(n) = na.

 
Example 3.7.2. (Linear transformations) Let V and W be vector spaces. Since any vector space is an abelian group under vector addition, any linear transformation between vector spaces is a group homomorphism.

 
Proposition 3.7.2. If µ : G1 -> G2 is a group homomorphism, then

(a) µ(e) = e;

(b) (µ(a))-1 = µ(a-1) for all a in G1;

(c) for any integer n and any a in G1, we have µ(an) = (µ(a))n;

(d) if a is in G1 and a has order n, then the order of µ(a) in G2 is a divisor of n.

 
Example 3.7.4. (Homomorphisms defined on cyclic groups) Let C be a cyclic group, denoted multiplicatively, with generator a. If µ : C -> G is any group homomorphism, and µ(a) = g, then the formula must hold. Since every element of C is of the form am for some integer m, this means that µ is completely determined by its value on a.
If C is infinite, then for an element g of any group G, the formula µ(am) = gm defines a homomorphism.
If |C|=n and g is any element of G whose order is a divisor of n, then the formula µ(am) = gm defines a homomorphism.

 
Example 3.7.5. (Homomorphisms from Zn to Zk) Any homomorphism µ : Zn -> Zk is completely determined by µ([1]n), and this must be an element [m]k of Zk whose order is a divisor of n. Then the formula µ([x]n) = [mx]k, for all [x]n in Zn, defines a homomorphism. Furthermore, every homomorphism from Zn into Zk must be of this form. The image µ(Zn) is the cyclic subgroup generated by [m]k.

 
Definition 3.7.3. Let µ : G1 -> G2 be a group homomorphism. Then

{ x in G1 | µ(x) = e }

is called the kernel of µ, and is denoted by ker(µ).

 
Proposition 3.7.4. Let µ : G1 -> G2 be a group homomorphism, with K = ker(µ).

(a) K is a normal subgroup of G.

(b) The homomorphism µ is one-to-one if and only if K = {e}.

 
Definition 3.7.5. A subgroup H of the group G is called a normal subgroup if

ghg-1 belongs to H

for all h in H and g in G.

 
Proposition 3.7.6. Let µ : G1 -> G2 be a group homomorphism.

(a) If H1 is a subgroup of G1, then µ(H1) is a subgroup of G2.
If µ is onto and H1 is normal in G1, then µ(H1) is normal in G2.

(b) If H2 is a subgroup of G2, then

µ-1 (H2) = { x in G1 | µ(x) is in H2 }

is a subgroup of G1.
If H2 is normal in G2, then µ-1(H2) is normal in G1.

 
Let µ : G1 -> G2 be a group homomorphism. The function µ determines an equivalence relation of G1 by setting a ~ b if µ(a) = µ(b). The notation G1/µ is used for the set of equivalence classes of this relation.

 
Proposition 3.7.7. Let µ : G1 -> G2 be a group homomorphism. Then multiplication of equivalence classes in G1/µ is well-defined, and G1/µ is a group under this multiplication. The natural mapping : G1 -> G1/µ defined by (x) = [x]µ is a group homomorphism.

 
Theorem 3.7.8. Let µ : G1 -> G2 be a group homomorphism. Then G1/µ is isomorphic to µ(G1).

 
Proposition 3.7.9. Let µ : G1 -> G2 be a group homomorphism, and let a,b belong to G1. The following conditions are equivalent:

(1) µ(a) = µ(b)

(2) ab-1 belongs to ker(µ);

(3) a = kb for some k in ker(µ);

(4) b-1a belongs to ker(µ);

(5) a = bk for some k in ker(µ);



§ 3.7 Homomorphisms: Solved problems

In Section 3.4 we introduced the concept of an isomorphism, and studied in detail what it means for two groups to be isomorphic. This section studies functions that respect the group operations but may not be one-to-one and onto. There are many important examples of group homomorphisms that are not isomorphisms, and, in fact, homomorphisms provide the way to relate one group to another. With the new terminology of this section, the usual way to show that two groups G1 and G2 are isomorphic is to define a homomorphism µ : G1 -> G2, and then show that µ is onto and has a trivial kernel.

Theorem 3.7.8 is the most important result, but Proposition 3.7.6 is also useful, since for any group homomorphism µ : G1 -> G2 it describes the connections between subgroups of G1 and subgroups of G2. Examples 3.7.4 and 3.7.5 are important, because they give a complete description of all group homomorphisms between two cyclic groups.


17. Find all group homomorphisms from Z4 into Z10.     Solution

18. (a) Find the formulas for all group homomorphisms from Z18 into Z30.
    (b) Choose one of the nonzero formulas in part (a), and for this formula find the kernel and image, and show how elements of the image correspond to cosets of the kernel.
    Solution

19. (a) Show that Z7× is cyclic, with generator [3]7.
    (b) Show that Z17× is cyclic, with generator [3]17.
    (c) Completely determine all group homomorphisms from Z17× into Z7×.
    Solution

20. Define µ : Z4 × Z6 -> Z4 × Z3 by

µ ([x]4,[y]6) = ([x+2y]4,[y]3).

    (a) Show that µ is a well-defined group homomorphism.
    (b) Find the kernel and image of µ, and apply the fundamental homomorphism theorem.
    Solution

21. Let n and m be positive integers, such that m is a divisor of n. Show that
µ : Zn× -> Zm× defined by

µ ( [x]n ) = [x]m,     for all [x]n in Zn×,

is a well-defined group homomorphism.     Solution

22. For the group homomorphism µ : Z36× -> Z12× defined by

µ ( [x]36 ) = [x]12,     for all [x]36 in Z36×,

find the kernel and image of µ, and apply the fundamental homomorphism theorem.     Solution

23. Let G, G1, and G2 be groups. Let µ1 : G -> G1 and µ2 : G -> G2 be group homomorphisms. Prove that
µ : G -> G1 × G2 defined by

µ (x) = (µ1 (x), µ2 (x)),     for all x in G,

is a well-defined group homomorphism.     Solution

24. Let p and q be different odd primes. Prove that Zpq× is isomorphic to the direct product Zp× × Zq× .     Solution


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