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By iterating all of these subgroups we can notice that there is an overlap between (15,3) and (5,3) namely {(18,36,54,72), (12,24)}<br> | By iterating all of these subgroups we can notice that there is an overlap between (15,3) and (5,3) namely {(18,36,54,72), (12,24)}<br> | ||
We can remedy this by ignoring the subgroup (5,3) so that it is not counted twice. Counting the remaining subgroups we get:<br> | We can remedy this by ignoring the subgroup (5,3) so that it is not counted twice. Counting the remaining subgroups we get:<br> | ||
| − | |(15,1)| + |(15,3)| = 14*1 + 14*2 = 42. | + | |(15,1)| + |(15,3)| = 14*1 + 14*2 = 42.<br> |
--[[User:Jniederh|Jniederh]] 00:45, 26 February 2009 (UTC) | --[[User:Jniederh|Jniederh]] 00:45, 26 February 2009 (UTC) | ||
Revision as of 20:45, 25 February 2009
For this problem you have to use theorem 8.1 in the book. It states that |(a,b)| = lcm(|a|,|b|). In this case we want to find the subgroups so that |(a,b)| = 15 = lcm(|a|,|b|). First we must find valid values for a and b so that lcm(|a|,|b|) = 15. The 3 subgroups of Z_90 x Z_36 that we can use are {(15,1), (15,3), (5,3)}. Now that we have the 3 subgroups we must find their corresponding elements and count them all.
(15,1) = {(6,12,18,...,84), (0)}, notice that this is just {(1*90/15, 2*90/15,...,14*90/15), (0)}
(15,3) = {(6,12,18,...,84), (12,24)}
(5,3) = {(18,36,54,72), (12,24)}, again notice this is {(1*90/5,...,4*90/5), (12,24)}
By iterating all of these subgroups we can notice that there is an overlap between (15,3) and (5,3) namely {(18,36,54,72), (12,24)}
We can remedy this by ignoring the subgroup (5,3) so that it is not counted twice. Counting the remaining subgroups we get:
|(15,1)| + |(15,3)| = 14*1 + 14*2 = 42.
--Jniederh 00:45, 26 February 2009 (UTC)
