College Algebra Exam Review 129

# College Algebra Exam Review 129 - 2.7 QUOTIENT GROUPS AND...

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Unformatted text preview: 2.7. QUOTIENT GROUPS AND HOMOMORPHISM THEOREMS 139 a normal subgroup N , there is a surjective homomorphism with N as ker- nel, and, on the other hand, a surjective homomorphism is essentially de- termined by its kernel. Theorem 2.7.6 also reveals the best way to understand a quotient group G=N . The best way is to find a natural model, namely some naturally defined group G together with a surjective homomorphism ' W G ! G with kernel N . Then, according to the theorem, G=N Š G . With this in mind, we take another look at the examples given above, as well as several more examples. Example 2.7.7. Let a be an element of order n in a group H . There is a homomorphism ' W Z ! H given by k 7! a k . This homomorphism has range h a i and kernel n Z . Therefore, by the homomorphism theorem, Z =n Z Š h a i . In particular, if D e 2 i=n , then '.k/ D k induces an isomorphism of Z =n Z onto the group C n of n th roots of unity in C ....
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