De¬nition 8.4. Let G be an abelian group, and let H be a non-empty subset

of G such that

(i) a + b ∈ H for all a, b ∈ H, and

(ii) ’a ∈ H for all a ∈ H.

Then H is called a subgroup of G.

In words: H is a subgroup of G if it is closed under the group operation

and taking inverses.

Multiplicative notation: if the abelian group G in the above de¬nition is

written using multiplicative notation, then H is a subgroup if ab ∈ H and

a’1 ∈ H for all a, b ∈ H.

Theorem 8.5. If G is an abelian group, and H is a subgroup of G, then

H contains 0G ; moreover, the binary operation of G, when restricted to H,

yields a binary operation that makes H into an abelian group whose identity

is 0G .

Proof. First, to see that 0G ∈ H, just pick any a ∈ H, and using both

properties of the de¬nition of a subgroup, we see that 0G = a + (’a) ∈ H.

Next, note that by property (i) of De¬nition 8.4, H is closed under ad-

dition, which means that the restriction of the binary operation “+” on G

to H induces a well de¬ned binary operation on H. So now it su¬ces to

show that H, together with this operation, satisfy the de¬ning properties

of an abelian group. Associativity and commutativity follow directly from

the corresponding properties for G. Since 0G acts as the identity on G, it

does so on H as well. Finally, property (ii) of De¬nition 8.4 guarantees that

every element a ∈ H has an inverse in H, namely, ’a. 2

Clearly, for an abelian group G, the subsets G and {0G } are subgroups.

These are not very interesting subgroups. An easy way to sometimes ¬nd

other, more interesting, subgroups within an abelian group is by using the

following two theorems.

Theorem 8.6. Let G be an abelian group, and let m be an integer. Then

mG := {ma : a ∈ G} is a subgroup of G.

Proof. For ma, mb ∈ mG, we have ma+mb = m(a+b) ∈ mG, and ’(ma) =

m(’a) ∈ mG. 2

186 Abelian groups

Theorem 8.7. Let G be an abelian group, and let m be an integer. Then

G{m} := {a ∈ G : ma = 0G } is a subgroup of G.

Proof. If ma = 0G and mb = 0G , then m(a + b) = ma + mb = 0G + 0G = 0G

and m(’a) = ’(ma) = ’0G = 0G . 2

Multiplicative notation: if the abelian group G in the above two theorems

is written using multiplicative notation, then we write the subgroup of the

¬rst theorem as Gm := {am : a ∈ G}. The subgroup in the second theorem

is denoted in the same way: G{m} := {a ∈ G : am = 1G }.

Example 8.20. For every integer m, the set mZ is the subgroup of the

additive group Z consisting of all integer multiples of m. Two such subgroups

mZ and m Z are equal if and only if m = ±m . The subgroup Z{m} is equal

to Z if m = 0, and is equal to {0} otherwise. 2

Example 8.21. Let n be a positive integer, let m ∈ Z, and consider the

subgroup mZn of the additive group Zn . Now, [b]n ∈ mZn if and only if

there exists x ∈ Z such that mx ≡ b (mod n). By Theorem 2.7, such an

x exists if and only if d | b, where d := gcd(m, n). Thus, mZn consists

precisely of the n/d distinct residue classes

[i · d]n (i = 0, . . . , n/d ’ 1),

and in particular, mZn = dZn .

Now consider the subgroup Zn {m} of Zn . The residue class [x]n is in

Zn {m} if and only if mx ≡ 0 (mod n). By Theorem 2.7, this happens if

and only if x ≡ 0 (mod n/d), where d = gcd(m, n) as above. Thus, Zn {m}

consists precisely of the d residue classes

[i · n/d]n (i = 0, . . . , d ’ 1),

and in particular, Zn {m} = Zn {d} = (n/d)Zn . 2

Example 8.22. For n = 15, consider again the table in Example 2.3. For

m = 1, 2, 3, 4, 5, 6, the elements appearing in the mth row of that table

form the subgroup mZn of Zn , and also the subgroup Zn {n/d}, where d :=

gcd(m, n). 2

Because the abelian groups Z and Zn are of such importance, it is a good

idea to completely characterize all subgroups of these abelian groups. As

the following two theorems show, the subgroups in the above examples are

the only subgroups of these groups.

8.2 Subgroups 187

Theorem 8.8. If G is a subgroup of Z, then there exists a unique non-

negative integer m such that G = mZ. Moreover, for two non-negative

integers m1 and m2 , we have m1 Z ⊆ m2 Z if and only if m2 | m1 .

Proof. Actually, we have already proven this. One only needs to observe

that a subset G of Z is a subgroup if and only if it is an ideal of Z, as

de¬ned in §1.2 (see Exercise 1.7). The ¬rst statement of the theorem then

follows from Theorem 1.5. The second statement follows easily from the

de¬nitions, as was observed in §1.2. 2

Theorem 8.9. If G is a subgroup of Zn , then there exists a unique positive

integer d dividing n such that G = dZn . Also, for positive divisors d1 , d2 of

n, we have d1 Zn ⊆ d2 Zn if and only if d2 | d1 .

Proof. Let ρ : Z ’ Zn be the map that sends a ∈ Z to [a]n ∈ Zn . Clearly, ρ

is surjective. Consider the pre-image ρ’1 (G) ⊆ Z of G.

We claim that ρ’1 (G) is a subgroup of Z. To see this, observe that for

a, b ∈ Z, if [a]n and [b]n belong to G, then so do [a + b]n = [a]n + [b]n and

’[a]n = [’a]n , and thus a + b and ’a belong to the pre-image.

Since ρ’1 (G) is a subgroup of Z, by the previous theorem, we have

ρ’1 (G) = dZ for some non-negative integer d. Moreover, it is clear that

n ∈ ρ’1 (G), and hence d | n. That proves the existence part of the theorem.

Next, we claim that for any divisor d of n, we have ρ’1 (dZn ) = dZ. To see

this, note that ρ’1 (dZn ) consists of all integers b such that dx ≡ b (mod n)

has an integer solution x, and by Theorem 2.7, this congruence admits a

solution if and only if d | b. That proves the claim.

Now consider any two positive divisors d1 , d2 of n. Since d1 Zn ⊆ d2 Zn

if and only if ρ’1 (d1 Zn ) ⊆ ρ’1 (d2 Zn ), the remaining statements of the

theorem follow from the corresponding statements of Theorem 8.8 and the

above claim. 2

Of course, not all abelian groups have such a simple subgroup structure.

Example 8.23. Consider the group G = Z2 — Z2 . For any non-zero ± ∈ G,

± + ± = 0G . From this, it is easy to see that the set H = {0G , ±} is a

subgroup of G. However, for any integer m, mG = G if m is odd, and

mG = {0G } if m is even. Thus, the subgroup H is not of the form mG for

any m. 2

Example 8.24. Consider again the group Z— , for n = 15, discussed in

n

Example 8.10. As discussed there, we have Z— = {[±1], [±2], [±4], [±7]}.

15

188 Abelian groups

Therefore, the elements of (Z— )2 are

15

[1]2 = [1], [2]2 = [4], [4]2 = [16] = [1], [7]2 = [49] = [4];

thus, (Z— )2 has order 2, consisting as it does of the two distinct elements

15

[1] and [4].

Going further, one sees that (Z— )4 = {[1]}. Thus, ±4 = [1] for all ± ∈ Z— .

15 15

By direct calculation, one can determine that (Z— )3 = Z— ; that is, cubing

15 15

—.

simply permutes Z15

For any integer m, write m = 4q + r, where 0 ¤ r < 4. Then for any

± ∈ Z— , we have ±m = ±4q+r = ±4q ±r = ±r . Thus, (Z— )m is either Z— ,

15 15 15

— )2 , or {[1]}.

(Z15

However, there are certainly other subgroups of Z— ” for example, the

15

subgroup {[±1]}. 2

Example 8.25. Consider again the group Z— from Example 8.11. As dis-

5

— = {[±1], [±2]}. Therefore, the elements of (Z— )2 are

cussed there, Z5 5