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The lcm can be equal 9:
 
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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.
 
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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).
 
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-K. Brumbaugh
 

Revision as of 14:19, 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)

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

Dr. Paul Garrett