# The following theorems while very simple to prove are

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The following theorems, while very simple to prove, are also very useful. Theorem 4.18 If H is a subgroup of an abelian group G , then the map f : G G/H given by f ( a ) = a + H is a surjective homomorphism whose kernel is H . This is sometimes called the “natural” map from G to G/H . Proof. Exercise. 2 Theorem 4.19 Let G and G 0 be abelian groups. Let f be a homomorphism from G into G 0 . Then the map ¯ f : G/ ker( f ) f ( G ) that sends the coset a + ker( f ) for a G to f ( a ) is unambiguously defined and is an isomorphism of G/ ker( f ) with f ( G ) . Proof. Exercise. 2 Theorem 4.20 Let G and G 0 be abelian groups. Let f be a homomorphism from G into G 0 . The subgroups of G containing ker( f ) are in one-to-one correspondence with the subgroups of f ( G ) , where the the subgroup H in G containing ker( f ) corresponds to the subgroup f ( H ) in f ( G ) . Proof. Exercise. 2 Theorem 4.21 Let G be an abelian group with subgroups H 1 , H 2 such that H 1 H 2 = { 0 G } . Then the map that sends ( h 1 , h 2 ) H 1 × H 2 to h 1 + h 2 H 1 + H 2 is an isomorphism of H 1 × H 2 with H 1 + H 2 . Proof. Exercise. 2 Example 4.27 For any abelian group G and any integer m , the map that sends a G to ma G is clearly a homomorphism from G into G . The image of this homomorphism is mG . We call this map the m -multiplication map on G . If G is written multiplicatively, we call this the m -power map on G . 2 Example 4.28 Consider the m -multiplication map on Z . The image of this map is m Z , and the kernel is { 0 } if m 6 = 0, and is Z if m = 0. 2 Example 4.29 Consider the m -multiplication map on Z n . The image of this map is m Z n , which as we saw above in Example 4.17 is a subgroup of Z n of order n/d , where d = gcd( n, m ). Thus, this map is bijective if and only if d = 1, in which case it is an isomorphism of Z n with itself. 2 Example 4.30 For positive integer n , consider the natural map f : Z Z /n Z = Z n . Theo- rem 4.20 says that this map gives a one-to-one correspondence between the subgroups of Z con- taining n Z and the subgroups of Z n . Moreover, it follows from Theorem 4.7 that the subgroups of Z containing n Z are precisely m Z for m | n . From this, it follows that the subgroups of Z n are precisely m Z n for m | n . We already proved this in Theorem 4.8. 2 Example 4.31 As was demonstrated in Example 4.25, the quotient group Z * 15 / ( Z * 15 ) 2 is isomorphic to Z 2 × Z 2 . 2 27

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Example 4.32 Let G 1 , G 2 be abelian groups. The map that sends ( a 1 , a 2 ) G 1 × G 2 to a 1 G 1 is a homomorphism from G 1 × G 2 to G 1 . Its image is G 1 , and its kernel is { 0 G 1 } × G 2 . 2 Example 4.33 If G = G 1 × G 2 for abelian groups G 1 and G 2 , and H 1 is a subgroup of G 1 and H 2 is a subgroup of G 2 , then H := H 1 × H 2 is a subgroup of G , and G/H = G 1 /H 1 × G 2 /H 2 . 2 4.5 Cyclic Groups Let G be an abelian group. For a G , define h a i := { za : z Z } . It is clear that h a i is a subgroup of G , and moreover, that any subgroup H of G that contains a must also contain h a i . The subgroup h a i is called the subgroup generated by a . Also, one defines the order of a to be the order of the subgroup h a i , which is denoted ord( a ).
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• Spring '13
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