(New page: Chapter 4: Problem 9 From theorem 4.3 we know that "for each positive divisor k of n, the group <a> has exactly one subgroup of order k" ==> <a^(n/k)> "For each positive divisor k of n, t...)
 
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Chapter 4: Problem 9 From theorem 4.3 we know that "for each positive divisor k of n, the group <a> has exactly one subgroup of order k" ==> <a^(n/k)>
 
  
"For each positive divisor k of n, the set <n/k> is the unique subgroup of Zn of order k. Most importantly these are the only subgroups of Zn."
 
 
Therefore if we follow the similiar examples on page 78 and 79 (example 5) we can see that for each divisor k of 20, the set <20/k> is the unique subgroup of Z20 of order k and are the only subgroups of Z20. Thus <1>, <2>, <4>, <5>, <10> and <20> are the subgroups of Z20.
 
 
Similarily, from the example on page 78 we see that <a>, <a^2> , <a^4>, <a^5>, <a^10>, <a^20> are the subgroups of G.
 
 
The generators are just the divisors of 20 (for Z20) or are a^k with k being a divisor of 20 (for the set G). ==> 1 , 2 , 4, 5, 10, 20 ==> a, a^2, a^4, a^5, a^10, a^20
 

Latest revision as of 03:23, 30 September 2012

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett