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## Homework Statement

Remember, the set of groups (G

_{n}, *), the group of multiplicatively-invertible elements of Z/n under multiplication. For p a prime, the elements of G

_{p}are all elements of Z/p except 0; for n not a prime, the elements of G

_{n}are all the elements of Z/n except 0 and those (besides 1) which divide n and all multiples of those elements.

a) Consider G

_{15}. What are the elements of G

_{15}, and what is the order of each elements? What group is G

_{15}isomorphic to?

## Homework Equations

Fermats theorem (maybe?) a

^{p-1}= 1 mod p

## The Attempt at a Solution

Ok I was having a little trouble understanding this at first but I think I'm starting to a little bit. For the first part, What are the elements of G

_{15}:

I think what the statement at the top is trying to tell me is that it is all of the elements of Z/15 except for 0 and the elements that divide 15 (besides 1) and the elements that are multiples of those elements.

So if that logic is correct I come up with the group: G

_{15}= {1, 2, 4, 7, 8, 11, 13, 14} Since 3 and 5 divide 15 and 6, 9, 12, and 10 are multiple of either 3 or 5 they are excluded from the group.

Next it asks for the order of each element, this is where I am a little confused. I found it using a calculator and just multiplying the elements and modding them until i get an even result that would bring it back to 1, but I'm thinking there is a more systematic way of going about this (maybe fermat's theorem?). Anyway I found that:

1 has order 1.

2 has order 4.

4 has order 2.

7 has order 4.

8 has order 4.

11 has order 2.

13 has order 4.

14 has order 2.

Now the last part asks what group G

_{15}is isomorphic to. I'm assuming it is isomorphic to Z/4 x Z/2 as both groups have order 8 with 4 elements of order 4, 3 of order 2 and 1 of order 1.

G

_{15}--> Z/4 x Z/2

1 --> e

2 --> a

4 --> a

^{2}

7 --> ab

8 --> a

^{3}

11 --> b

13 --> a

^{3}b

14 -->a

^{2}b

Upon checking, these mappings hold. Again I had to use a calculator to check all of this and I'm unsure if there is an easier way to do this.