Finding elements of G_15 and their orders.

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Homework Statement


Remember, the set of groups (Gn, *), the group of multiplicatively-invertible elements of Z/n under multiplication. For p a prime, the elements of Gp are all elements of Z/p except 0; for n not a prime, the elements of Gn 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?) ap-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.
 

Answers and Replies

  • #2
22,089
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Seems fine. At this point I cant 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.
 
  • #3
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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.
 
  • #4
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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?
 
Last edited:
  • #5
Dick
Science Advisor
Homework Helper
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Looks ok to me.
 

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