# Solutions to Problem Set 11

Math 345/504

2-26-2018

1. (a) List the elements of .

(b) Compute in .

(c) Compute in .

(d) List the elements of the cyclic subgroup in .

(a)

(b)

(c) Since , it follows that .

(d)

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.

[MATH 504]

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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