1. (a) List the elements of .
(b) Compute in .
(c) Compute in .
(d) List the elements of the cyclic subgroup in .
(c) Since , it follows that .
2. Prove directly that is not cyclic.
I'll show that if , then there is some rational number that is not contained in . This will prove that , and hence that is not cyclic.
First, if , then . Any nonzero rational number is not contained in .
Now assume .
x and and both elements of , and since one of them must be positive. Without loss of generality, assume .
I claim that . Suppose to the contrary that . Then for some ,
Since , I get , contradicting the fact that n is na integer.
Therefore, is not cyclic.
3. This is a multiplication table for :
Find all the generators of .
Note that and . Hence, 5 does not have order 1, 2, or 3. Hence, 5 must have order 6, and it's a genrator.
Thus, is cyclic of order 6. The generators in are 1 and 5. So the generators in are and .
We have to be watchful in case we goof up, but we are uncertain as to what we are going to goof up. - Ch\"ogyam Trungpa
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