Cyclic Subgroups of P15: Homework Solutions

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Homework Statement


Consider the set P15 of all integer numbers less than 15 that are mutually prime with 15: P15 = {1, 2, 4, 7, 8, 11, 13, 14}. It is a group under multiplication modulo 15.

(a) P15 has six cyclic groups. Find them.
my answer: <3>=<6>=<9>=<12>= {0, 3, 6 , 9, 12} and <5>=<10>= {0, 5, 10}

(b) For each cyclic subgroup of order 4 give an isomorphism with Z_4.
Well, at this point I figure I must have done (a) wrong since I do not have any subgroups with order 4. If I did I would know how to give an isomorphism with Z_4 so that is not a problem.

(c) Find a noncyclic subgroup of order 4 in P15.
I thought P15 was cyclic and a subgroup of a cyclic group is cyclic, right?

(d) To what well known group is (c) isomorphic?
Isn't this the same question as (b)?

(e) Why can we be sure that P15 has no other noncyclic subgroups of order 4?

(f) Is P15 cyclic?
I thought so but that makes some of the other questions irrelevant.
 
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Let's just start with the first one. Z15 is cyclic and contains 0. P15 as you've defined it doesn't contain 0. 0 isn't an integer that's mutually prime with 15. Try and work on that one first.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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