Finding elements of G_15 and their orders.

[nhartung]

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 G₁₅. What are the elements of G₁₅, and what is the order of each elements? What group is G₁₅ 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 G₁₅:

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: G₁₅ = {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 G₁₅ 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.

G₁₅ --> Z/4 x Z/2

1 --> e
2 --> a
4 --> a²
7 --> ab
8 --> a³
11 --> b
13 --> a³b
14 -->a²b

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.

 

[micromass]

Seems fine. At this point I can't see any other method then a brute force calculation.

One thing that can be shortened is to find the group isomorphic to G_15. You don't need to construct the map for that and to check that it's a homomorphism. It can be shorter like this (and it avoids all the calculations):

G_15 is an abelian group of order 8. Through the fundamental theorem of abelian groups, the only abelian groups of order 8 are Z8, Z4xZ2, Z2xZ2xZ2.
But G_15 is not Z8, since it contains no elements of order 8.
Similarly, G15 is not Z2xZ2xZ2, since G_15 contains elements of order 4 and Z2xZ2xZ2 doesnt.

So the only possibility is that G_15 is actually Z4xZ2.

 

[nhartung]

Ah, I like that proof a lot better. Thanks for the quick response, if I have any questions of the other parts of this question (didn't post them yet) i'll post them here.

 

[nhartung]

ok part c asks: Demonstrate that {1,11} in G_15 is a subgroup of G_15, and {1,11} in G_30 is a subgroup of G_30. Find the cosets of {1,11} in each of the groups.

Part b had me find G_30 and if it was isomorphic to G_15. From that I found G_30 = {1, 7, 11, 13, 17, 19, 23, 29} which is also isomorphic to Z/4 x Z/2 so these groups are isomorphisms to each other.

So for part c these are both obvious subgroups having closure(11 is of order 2), identity (1), and inverses (both are their own inverse). Now I need to find the cosets of these 2 subgroups here's what I've come up with:

for G_15: H = {1, 11}
2*H = {2, 7}
4*H = {4, 14}
8*H = {8, 13}

for G_30: H = {1, 11}
7*H = {7, 17}
13*H = {13, 23}
19*H = {19,29}

Maybe I should write it as 2*H = H*2 = {2, 7} showing that the group is abelian and an action on either side gives the same result?

These are the only distinct cosets of H in both cases.

Does this look correct?

 

[Dick]

Looks ok to me.