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[[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) | ||
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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, meaning that there is a representative of orders 1, 3 and 9 present. | ||
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| + | ex. a = 8 (|4| = 9), b = 0 (|0| = 1), c = 5 (|5| = 3) => H = (8,0,5), under addition, 9*H = (72,0,45) = (0,0,0). | ||
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| + | -K. Brumbaugh | ||
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| + | ---- | ||
| + | [[Category:MA453Spring2009Walther]] | ||
| + | Isn't |4|=3 in Z12, same with |8|? I don't see how you can find an element of order 9 in Z12.-Paul | ||
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| + | -------- | ||
| + | We are looking for a subgroup of order 9 implying the subgroup has 9 elements.-Rob | ||
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| + | Rob is correct, we are looking for a subgroup of order 9, not an element of order 9. Since there is no limitation on the subgroup (ie it has to be normal, or something) I believe any 9 elements from the group are fine...--[[User:Bcaulkin|Bcaulkin]] 23:14, 25 February 2009 (UTC) | ||
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| + | [[Category:MA453Spring2009Walther]] | ||
| + | Why would the subgroup have to be normal? --[[User:Lmiddlet|Lmiddlet]] 00:34, 26 February 2009 (UTC) | ||
Latest revision as of 19:34, 25 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, meaning that there is a representative of orders 1, 3 and 9 present.
ex. a = 8 (|4| = 9), b = 0 (|0| = 1), c = 5 (|5| = 3) => H = (8,0,5), under addition, 9*H = (72,0,45) = (0,0,0).
-K. Brumbaugh
Isn't |4|=3 in Z12, same with |8|? I don't see how you can find an element of order 9 in Z12.-Paul
We are looking for a subgroup of order 9 implying the subgroup has 9 elements.-Rob
Rob is correct, we are looking for a subgroup of order 9, not an element of order 9. Since there is no limitation on the subgroup (ie it has to be normal, or something) I believe any 9 elements from the group are fine...--Bcaulkin 23:14, 25 February 2009 (UTC)
Why would the subgroup have to be normal? --Lmiddlet 00:34, 26 February 2009 (UTC)
