# The fact that this definition is unambiguous follows

• Notes
• 74

This preview shows pages 30–33. Sign up to view the full content.

The fact that this definition is unambiguous follows immediately from Theorem 4.14. Also, one can easily verify that this operation defines an abelian group. The resulting group is called the quotient group of G modulo H , and is denoted G/H . The order of the group G/H is sometimes denoted [ G : H ] and is called the index of H in G . If G is of finite order, then by Theorem 4.12, [ G : H ] = | G | / | H | . Multiplicative notation: if G is written multiplicatively, then the definition of the group opera- tion of G/H is expressed ( aH ) · ( bH ) := ( ab ) H. Theorem 4.15 If H H 0 are subgroups of an abelian group G , and [ G : H ] is finite, then [ G : H ] = [ G : H 0 ] · [ H 0 : H ] . Proof. Exercise. 2 25

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

Example 4.24 For the additive group of integers Z and the subgroup n Z for n > 0, the quotient group Z /n Z is precisely the same as the additive group Z n that we have already defined. For n = 0, Z /n Z is essentially just a “renaming” of Z . 2 Example 4.25 Let us return to Example 4.20. The multiplicative group Z * 15 , as we saw, is of order 8. The subgroup ( Z * 15 ) 2 has order 2. Therefore, the quotient group has order 4. Indeed, the cosets are α 00 = { [1] , [4] } , α 01 = { [ - 1] , [ - 4] } , α 10 = { [2] , [ - 7] } , and α 11 = { [7] , [ - 2] } . In the group Z * 15 / ( Z * 15 ) 2 , α 00 is the identity; moreover, we have α 2 01 = α 2 10 = α 2 11 = α 00 and α 01 α 10 = α 11 , α 10 α 11 = α 01 , α 10 α 11 = α 01 . This completely describes the behavior of the group operation of the quotient group. Note that this group is essentially just a “renaming” of the group Z 2 × Z 2 . 2 Example 4.26 As we saw in Example 4.21, ( Z * 5 ) 2 = { [ ± 1] } . Therefore, the quotient group Z * 5 / ( Z * 5 ) 2 has order 2. The cosets of ( Z * 5 ) 2 in Z * 5 are { [ ± 1] } and { [ ± 2 } . 2 4.4 Group Homomorphisms and Isomorphisms Definition 4.16 A homomorphism from an abelian group G to an abelian group G 0 is a function f : G G such that f ( a + b ) = f ( a ) + f ( b ) for all a, b G . The set f - 1 (1 G 0 ) is called the kernel of f , and is denoted ker( f ) . The set f ( G ) is called the image of f . If f is bijective, then f is called an isomorphism of G with G 0 . It is easy to see that if f is an isomorphism of G with G 0 , then the inverse function f - 1 is an isomorphism of G 0 with G . If such an isomorphism exists, we say that G and G 0 are isomorphic , and write G = G 0 . we stress that an isomorphism of G with G 0 is essentially just a “renaming” of the group elements — all structural properties of the group are preserved. Theorem 4.17 Let f be a homomorphism from an abelian group G to an abelian group G 0 . 1. f (0 G ) = 0 G 0 . 2. f ( - a ) = - f ( a ) for all a G . 3. f ( na ) = nf ( a ) for all n Z and a G . 4. For any subgroup H of G , f ( H ) is a subgroup of G 0 . 5. ker( f ) is a subgroup of G . 6. For all a, b G , f ( a ) = f ( b ) if and only if a b (mod ker( f )) . 7. f is injective if and only if ker( f ) = { 0 G } . 8. For any subgroup H 0 of G 0 , f - 1 ( H 0 ) is a subgroup of G containing ker( f ) . 9. For any subgroup H of G , f - 1 ( f ( H )) = H + ker( f ) . 26
Proof. Exercise. 2 Part (7) of the above theorem is particular useful: to check that a homomorphism is injective, it suffices to determine if ker( f ) = { 0 G } .

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

This is the end of the preview. Sign up to access the rest of the document.
• Spring '13
• MRR

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern