Line 8: Line 8:
 
[[Category:MA453Spring2009Walther]]
 
[[Category:MA453Spring2009Walther]]
 
I don't think that there is a subgroup of order 9. I think because the lcm of the orders can not be equal to 9, therefore, there is no subgroup of order 9. --[[User:Awika|Awika]] 12:39, 24 February 2009 (UTC)
 
I don't think that there is a subgroup of order 9. I think because the lcm of the orders can not be equal to 9, therefore, there is no subgroup of order 9. --[[User:Awika|Awika]] 12:39, 24 February 2009 (UTC)
 +
 +
----
 +
The lcm can be equal 9:
 +
 +
The subgroups have the form (a,b,c) = H. a is in Z12, b is in Z4 and c is in Z15. By theorem 8.1 in order for |H| = 9, lcm(|a|,|b|,|c|) = 9, thus |a|, |b|, |c| = 1, 3 or 9, who's lcm is 9. Thus, find a value of a, b and c that this statement is true.
 +
 +
ex. a = 4 (|4| = 3), b = 0 (|0| = 1), c = 5 (|5| = 3) => H = (4,0,5), under addition, 9*H = (36,0,45) = (0,0,0).
 +
 +
-K. Brumbaugh

Revision as of 14:18, 24 February 2009


Can anyone explain how to do this problem. I understand that since this has an order of 9 then it is generated by 6, but I just don't know how to find the subgroups. Thanks! --Lchinn 22:03, 23 February 2009 (UTC)



I don't think that there is a subgroup of order 9. I think because the lcm of the orders can not be equal to 9, therefore, there is no subgroup of order 9. --Awika 12:39, 24 February 2009 (UTC)


The lcm can be equal 9:

The subgroups have the form (a,b,c) = H. a is in Z12, b is in Z4 and c is in Z15. By theorem 8.1 in order for |H| = 9, lcm(|a|,|b|,|c|) = 9, thus |a|, |b|, |c| = 1, 3 or 9, who's lcm is 9. Thus, find a value of a, b and c that this statement is true.

ex. a = 4 (|4| = 3), b = 0 (|0| = 1), c = 5 (|5| = 3) => H = (4,0,5), under addition, 9*H = (36,0,45) = (0,0,0).

-K. Brumbaugh

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett