[1]2 = [1], [2]2 = [4] = [’1];

thus, (Z— )2 = {[±1]} and has order 2.

5

There are in fact no other subgroups of Z— besides Z— , {[±1]}, and {[1]}.

5 5

Indeed, if H is a subgroup containing [2], then we must have H = Z— : 5

2 = [4] = [’1] ∈ H, which implies [’2] ∈ H as well. The

[2] ∈ H implies [2]

same holds if H is a subgroup containing [’2]. 2

Example 8.26. Consider again the group of arithmetic functions f , such

that f (1) = 0, with multiplication de¬ned by the Dirichlet product, dis-

cussed in Example 8.13. By the results of Exercises 2.21 and 2.28, we see

that the subset of all multiplicative arithmetic functions is a subgroup of

this group. 2

The following two theorems may be used to simplify verifying that a subset

is a subgroup.

Theorem 8.10. If G is an abelian group, and H is a non-empty subset of

G such that a ’ b ∈ H for all a, b ∈ H, then H is a subgroup of G.

Proof. Since H is non-empty, let c be an arbitrary element of H. Then

0G = c ’ c ∈ H. It follows that for all a ∈ H, we have ’a = 0G ’ a ∈ H,

and for all a, b ∈ H, we have a + b = a ’ (’b) ∈ H. 2

Theorem 8.11. If G is an abelian group, and H is a non-empty, ¬nite

subset of G such that a + b ∈ H for all a, b ∈ H, then H is a subgroup of G.

8.2 Subgroups 189

Proof. We only need to show that ’a ∈ H for all a ∈ H. Let a ∈ H be

given. If a = 0G , then clearly ’a = 0G ∈ H, so assume that a = 0G , and

consider the set S of all elements of G of the form ma, for m = 1, 2, . . . . Since

H is closed under addition, it follows that S ⊆ H. Moreover, since H is

¬nite, S must be ¬nite, and hence there must exist integers m1 , m2 such that

m1 > m2 > 0 and m1 a = m2 a; that is, ra = 0G , where r := m1 ’m2 > 0. We

may further assume that r > 1, since otherwise a = 0G , and we are assuming

that a = 0G . It follows that a + (r ’ 1)a = 0G , and so ’a = (r ’ 1)a ∈ S. 2

We close this section with two theorems that provide useful ways to build

new subgroups out of old subgroups.

Theorem 8.12. If H1 and H2 are subgroups of an abelian group G, then

so is

H1 + H2 := {h1 + h2 : h1 ∈ H1 , h2 ∈ H2 }.

Proof. Consider two elements in H1 + H2 , which we can write as h1 + h2 and

h1 + h2 , where h1 , h1 ∈ H1 and h2 , h2 ∈ H2 . Then by the closure properties

of subgroups, h1 +h1 ∈ H1 and h2 +h2 ∈ H2 , and hence (h1 +h2 )+(h1 +h2 ) =

(h1 + h1 ) + (h2 + h2 ) ∈ H1 + H2 . Similarly, ’(h1 + h2 ) = (’h1 ) + (’h2 ) ∈

H1 + H 2 . 2

Multiplicative notation: if the abelian group G in the above theorem is

written multiplicatively, then the subgroup de¬ned in the theorem is written

H1 · H2 := {h1 h2 : h1 ∈ H1 , h2 ∈ H2 }.

Theorem 8.13. If H1 and H2 are subgroups of an abelian group G, then

so is H1 © H2 .

Proof. If h ∈ H1 © H2 and h ∈ H1 © H2 , then since h, h ∈ H1 , we have

h + h ∈ H1 , and since h, h ∈ H2 , we have h + h ∈ H2 ; therefore, h + h ∈

H1 © H2 . Similarly, ’h ∈ H2 and ’h ∈ H2 , and therefore, ’h ∈ H1 © H2 .

2

Exercise 8.2. Show that if H is a subgroup of an abelian group G, then a

set H ⊆ H is a subgroup of G if and only if H is a subgroup of H .

Exercise 8.3. Let G be an abelian group with subgroups H1 and H2 . Show

that any subgroup H of G that contains H1 ∪ H2 contains H1 + H2 , and

H1 ⊆ H2 if and only if H1 + H2 = H2 .

Exercise 8.4. Let H1 be a subgroup of an abelian group G1 and H2 a

subgroup of an abelian group G2 . Show that H1 — H2 is a subgroup of

G1 — G2 .

190 Abelian groups

Exercise 8.5. Let G1 and G2 be abelian groups, and let H be a subgroup

of G1 — G2 . De¬ne

H1 := {h1 ∈ G1 : (h1 , h2 ) ∈ H for some h2 ∈ G2 }.

Show that H1 is a subgroup of G1 .

Exercise 8.6. Give an example of speci¬c abelian groups G1 and G2 , along

with a subgroup H of G1 — G2 , such that H cannot be written as H1 — H2 ,

where H1 is a subgroup of G1 and H2 is a subgroup of G2 .

8.3 Cosets and quotient groups

We now generalize the notion of a congruence relation.

Let G be an abelian group, and let H be a subgroup of G. For a, b ∈ G,

we write a ≡ b (mod H) if a ’ b ∈ H. In other words, a ≡ b (mod H) if and

only if a = b + h for some h ∈ H.

Analogously to Theorem 2.2, if we view the subgroup H as ¬xed, then

the following theorem says that the binary relation “· ≡ · (mod H)” is an

equivalence relation on the set G:

Theorem 8.14. Let G be an abelian group and H a subgroup of G. For all

a, b, c ∈ G, we have:

(i) a ≡ a (mod H);

(ii) a ≡ b (mod H) implies b ≡ a (mod H);

(iii) a ≡ b (mod H) and b ≡ c (mod H) implies a ≡ c (mod H).

Proof. For (i), observe that H contains 0G = a ’ a. For (ii), observe that if

H contains a ’ b, then it also contains ’(a ’ b) = b ’ a. For (iii), observe

that if H contains a’b and b’c, then it also contains (a’b)+(b’c) = a’c.

2

Since the binary relation “· ≡ · (mod H)” is an equivalence relation, it

partitions G into equivalence classes. It is easy to see (verify) that for any

a ∈ G, the equivalence class containing a is precisely the set a+H := {a+h :

h ∈ H}, and this set is called the coset of H in G containing a, and an

element of such a coset is called a representative of the coset.

Multiplicative notation: if G is written multiplicatively, then a ≡

b (mod H) means a/b ∈ H, and the coset of H in G containing a is

aH := {ah : h ∈ H}.

Example 8.27. Let G := Z and H := nZ for some positive integer n. Then

8.3 Cosets and quotient groups 191

a ≡ b (mod H) if and only if a ≡ b (mod n). The coset a + H is exactly the

same thing as the residue class [a]n . 2

Example 8.28. Let G := Z4 and let H be the subgroup 2G = {[0], [2]} of

G. The coset of H containing [1] is {[1], [3]}. These are all the cosets of H

in G. 2

Theorem 8.15. Any two cosets of a subgroup H in an abelian group G

have equal cardinality; that is, there is a bijective map from one coset to the

other.

Proof. It su¬ces to exhibit a bijection between H and a + H for any a ∈ G.

The map fa : H ’ a + H that sends h ∈ H to a + h is easily seen to be just

such a bijection. 2

An incredibly useful consequence of the above theorem is:

Theorem 8.16 (Lagrange™s theorem). If G is a ¬nite abelian group, and

H is a subgroup of G, then the order of H divides the order of G.

Proof. This is an immediate consequence of the previous theorem, and the

fact that the cosets of H in G partition G. 2

Analogous to Theorem 2.3, we have:

Theorem 8.17. Let G be an abelian group and H a subgroup. For

a, a , b, b ∈ G, if a ≡ a (mod H) and b ≡ b (mod H), then a + b ≡

a + b (mod H).

Proof. Now, a ≡ a (mod H) and b ≡ b (mod H) means that a = a+h1 and

b = b + h2 for h1 , h2 ∈ H. Therefore, a + b = (a + h1 ) + (b + h2 ) = (a + b) +

(h1 + h2 ), and since h1 + h2 ∈ H, this means that a + b ≡ a + b (mod H).

2

Let G be an abelian group and H a subgroup. Theorem 8.17 allows us

to de¬ne a binary operation on the collection of cosets of H in G in the

following natural way: for a, b ∈ G, de¬ne

(a + H) + (b + H) := (a + b) + H.

The fact that this de¬nition is unambiguous follows immediately from The-

orem 8.17. Also, one can easily verify that this operation de¬nes an abelian

group, where H acts as the identity element, and the inverse of a coset a+H

is (’a) + H. The resulting group is called the quotient group of G mod-

ulo H, and is denoted G/H. The order of the group G/H is sometimes

denoted [G : H] and is called the index of H in G.

192 Abelian groups

Multiplicative notation: if G is written multiplicatively, then the de¬nition

of the group operation of G/H is expressed

(aH) · (bH) := (ab)H.