The following theorems while very simple to prove are

Info icon This preview shows pages 32–34. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 32

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 ).
Image of page 33
Image of page 34
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern