Difference between revisions of "Chapter 8- Problem 34" - Rhea

Revision as of 20:19, 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)

Any cyclic subgroup of order 15 has $ \scriptstyle\phi(15)\ =\ $ 8 elements of order 15, so the number of elements of order 15 in $ \scriptstyle Z_{90}\oplus Z_{36} $ must be divisible by 8.

I found the number of order 15 elements by looking at these mutually exclusive cases:

Case 1: |a| = 15, |b| = 1 or 3

According to the Euler totient function, there are 8 choices for a, and 1+2 = 3 choices for b, giving a total of 8*3 = 24 elements in this case.

Case 2: |a| = 5, |b| = 3

Here we have 4 choices for a and 2 choices for b, so in total 8 elements.

So there are 24+8 = 32 elements of order 15 in $ \scriptstyle Z_{90}\oplus Z_{36} $. By Theorem 4.4 every cyclic subgroup of order 15 has 8 unique generators, that is, 8 elements of order 15 unique to that subgroup. So, there are 32/8 = 4 cyclic subgroups of order 15 in $ \scriptstyle Z_{90}\oplus Z_{36} $.

--Nick Rupley 01:19, 26 February 2009 (UTC)