# 2 22 because the abelian groups z and z n are of such

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2 22

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Because the abelian groups Z and Z n 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 sub- groups in the above examples are the only subgroups of these groups. Theorem 4.7 If G is a subgroup of Z , then there exists a unique non-negative integer m such that G = m Z . Proof. Actually, we have already proven this. One only needs to observe that a subset G is a subgroup if and only if it is an ideal (as defined in § 1.2), and then apply Theorem 1.4. 2 Theorem 4.8 If G is a subgroup of Z n , then there exists a unique positive integer m dividing n such that G = m Z n . Proof. Let G be a subgroup of Z n . Define G 0 := { a Z : [ a ] G } . It is easy to see that G = { [ a ] : a G 0 } . First, we claim that G 0 is a subgroup of Z . Suppose that a, b G 0 . This means that [ a ] G and [ b ] G , which implies that [ a + b ] = [ a ] + [ b ] G , and hence a + b G 0 . Similarly, if [ a ] G , then [ - a ] = - [ a ] G , and hence - a G 0 . By the previous theorem, it follows that G 0 is of the form m Z for some non-negative integer m . Moreover, note that n G 0 , since [ n ] = [0] is the identity element of Z n , and hence belongs to G . Therefore, m | n . So we have G = { [ a ] : a m Z } = m Z n . From the observations in Example 4.17, the uniqueness of m is clear. 2 Of course, not all abelian groups have such a simple subgroup structure. Example 4.18 Consider the group G = Z 2 × Z 2 . For any non-zero α G , α + α = 0 G . From this, it is easy to see that the set H = { 0 G , α } is a subgroup of G . However, for any integer m , mG = G if m is odd, and mG = { 0 G } if m is even. Thus, the subgroup H is not of the form mG for any m . 2 Example 4.19 Consider the group Z * n discussed in Example 4.9. The subgroup ( Z * n ) 2 plays an important role in some situations. Integers a such that [ a ] ( Z * n ) 2 are called quadratic residues modulo n . 2 Example 4.20 Consider again the group Z * n , for n = 15, discussed in Example 4.10. As discussed there, we have Z * 15 = { [ ± 1] , [ ± 2] , [ ± 4] , [ ± 7] } . Therefore, the elements of ( Z * 15 ) 2 are [1] 2 = [1] , [2] 2 = [4] , [4] 2 = [16] = [1] , [7] 2 = [49] = [4]; thus, ( Z * 15 ) 2 has order 2, consisting as it does of the two distinct elements [1] and [4]. Going further, one sees that ( Z * 15 ) 4 = { [1] } . Thus, α 4 = [1] for all α Z * 15 . By direct calculation, one can determine that ( Z * 15 ) 3 = Z * 15 ; that is, cubing simply permutes Z * 15 . For any integer m , write m = 4 q + r , where 0 r < 4. Then for any α Z * 15 , we have α m = α 4 q + r = α 4 q α r = α r . Thus, ( Z * 15 ) m is either Z * 15 , ( Z * 15 ) 2 , or { [1] } . However, there are certainly other subgroups of Z * 15 — for example, the subgroup { [ ± 1] } . 2 23
Example 4.21 Consider again the group Z * 5 from Example 4.11. As discussed there, Z * 5 = { [ ± 1] , [ ± 2] } . Therefore, the elements of ( Z * 5 ) 2 are [1] 2 = [1] , [2] 2 = [4] = [ - 1]; thus, ( Z * 5 ) 2 = { [ ± 1] } and has order 2. There are in fact no other subgroups of Z * 5 besides Z * 5 , { [ ± 1]] } , and { [1] } . Indeed, if H is a subgroup containing [2], then we must have H = Z * 5 : [2] H implies [2] 2 = [4] = [ - 1] H , which implies [ - 2] H as well. The same holds if H is a subgroup containing [ - 2]. 2 If G is an abelian group, and H 1 and H 2 are subgroups, we define H 1 + H 2 := { h 1 + h 2 : h 1 H 1 , h 2 H 2 } . Note that H 1 + H 2 contains H 1 H 2 .

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