# Proof exercise 2 if g 1 g k are abelian groups we can

• Notes
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Proof. Exercise. 2 If G 1 , . . . , G k are abelian groups, we can form the direct product G 1 ×· · ·× G k , which consists of all k -tuples ( a 1 , . . . , a k ) for a 1 G 1 , . . . , a k G k . We can view G 1 × · · · × G k in a natural way as an abelian group if we define the group operation “component wise”: ( a 1 , . . . , a k ) + ( b 1 , . . . , b k ) := ( a 1 + b 1 , . . . , a k + b k ) . 21

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Of course, the groups G 1 , . . . , G k may be different, and the group operation applied in the i th component corresponds to the group operation associated with G i . We leave it to the reader to verify that G 1 × · · · × G k is in fact an abelian group. In these notes, we have chosen only to discuss the notion of an abelian group. There is a more general notion of a group , which may be defined simply by dropping the commutivity condition in Definition 4.1, but we shall not need this notion in these notes, and restricting to abelian groups helps to simplify the discussion significantly. Nevertheless, many of the notions and results we discuss here regarding abelian groups extend (sometimes with slight modification) to general groups. Example 4.15 The set of 2 × 2 integer matrices with determinant ± 1 with respect to matrix multiplication forms a group, but not an abelian group. 2 4.2 Subgroups We next introduce the notion of a subgroup. Definition 4.4 Let G be an abelian group, and let H be a non-empty subset of G such that for all a, b H , a + b H , and for all a H , - a H . Then H is called a subgroup of G . Theorem 4.5 If G is an abelian group, and H is a subgroup, then the binary operation of G defines a binary operation on H , and with respect to this binary operation, H forms an abelian group whose identity is the same as that of G . Proof. Exercise. 2 Clearly, for an abelian group G , the subsets G and { 0 G } are subgroups. An easy way to find other, more interesting, subgroups within an abelian group is by using the following theorem: Theorem 4.6 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 Multiplicative notation: if the abelian group G in the above theorem is written using multiplica- tive notation, then we write the subgroup of that theorem G m := { a m : a G } . Example 4.16 For every integer m , the set m Z is a subgroup of the group Z . 2 Example 4.17 Let n be a positive integer, and let m Z . By the above theorem, m Z n is a subgroup of Z n ; however, we wish to give an explicit description of this subgroup. Consider a fixed residue class [ a ] for a Z . Now, [ a ] m Z n if and only if there exists x Z such that mx a (mod n ). By Theorem 2.5, such an x exists if and only if d | a , where d = gcd( m, n ). Thus, m Z n = d Z n , and consists precisely of the n/d distinct residue classes [ i · d ] ( i = 0 , . . . , n/d - 1) .
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• Spring '13
• MRR

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