Homework Help: Finding a group isomorphism under modulo multiplication

1. Nov 25, 2011

Hugheberdt

1. The problem statement, all variables and given/known data
I am given the group (G,$\bullet$) consisting of all elements that are invertible in the ring $Z$/20Z. I am to find the direct product of cyclic groups, which this group G is supposedly isomorphic to. I am also to describe the isomorphism.

2. Relevant equations

3. The attempt at a solution
I have found the elements of the group to be (1 3 7 9 11 13 17 19) and have set up a group table (which I am unfortunately unable to show here). Since I observed the group table too be seemingly divided into four squares size 4x4, the top left and bottom right being:

--|1 3 7 9
------------
1 |1 3 7 9
3 |3 9 1 7
7 |7 1 9 3
9 |9 7 3 9

and the lower left and upper right identical, but all elements added 10.

I thought I'd start by investigating the smaller group above and see if I can find some isomorphism from that group (let's call it H) to some cyclic group (a tempting candidate C$_{4}$ since the order of both groups is 4).

What I've learned about cyclic groups of order m is that they can be generated by the powers up to m-1 of some generator element x. In order to show that a group H is isomorphic to some cyclic group of order m you need to find an isomorphism β from H,* to that cyclic group. Specifically, you need to show that β(a*b)=β(a)*β(b) under the group operation * in H (for any two elements a and b in H), where β is a bijection.

I know a few things about isomorphisms; that the order of the isomorphs are equal, that identity maps to identity, that symmetry is preserved, that orders of elements in H are equal to the orders of their maps under β, etc.

I found that the map β(a(mod 5))=x^(a(mod5)) defines an isomorphism from H to K (a group of order 4), where x^1 is the identity element in K. Their group tables seem identical, apart from names of the elements. So far so good, I've managed to find a bijection from H to K, where the elements are powers of x (x,x^2,x^3,x^4). However, to have the group consist of 1,x,x^2,x^3 I'd like to lower the powers, which can be done with the bijection β$_{2}$: K to L , β$_{2}$(x^n)=x^n-1 (my plan is then to make a composition of functions). Here I stumble upon a few problems:

1. Does my first bijection β really obey β(a*b)=β(a)*β(b)?
For instance β(a(mod 5)$\bullet$b(mod 5))=x^a(mod 5)$\bullet$b(mod 5) = (x^a(mod 5))b(mod 5) = β(a(mod 5))*β(b(mod 5)) ????

2. If I replace x^n in the group table of G with x^(n-1) the group table pattern remains identical. If I on the other hand only replace the group elements and compute the new group table afresh, inconsistencies arise. Most notably, β(1*x^n)= (x^0(mod 5))n(mod 5)=1^n=1, which yields several identity elements. If I on the other hand, choose the ordinary addition operation for this new group L, I get a group table with a different pattern than the group tables for H and K.

3. In either case, since β(1)=x, β(3)=x^3, β(7)=x^2, β(9)=x^4, the elements get arranged in the "wrong order" (x,x^3,x^2,x^4 as opposed to the desired x,x^2,x^3,x^4). I thought of perhaps permuting the middle elements, but that would further mess up the group table.

I suspect that I'm getting several things wrong, so please correct me. I've searched the internet, these forums and my course litterature, but I don't feel that I find the answers to my questions.

Great thanks to anyone who chooses to help me out with this rather long query!

2. Nov 25, 2011

Dick

You are probably making this way too complicated. Your group has order 8. How many different ways are there to get a product of cyclic groups to have order 8? I would figure out which one it is by looking at orders of your group elements. The element 3 has order 4, right? In fact, it generates the order 4 cyclic group you've already found. What about the other elements? Try thinking about it that way for a bit. To find the explicit isomorphism just show how the generators of your group map to the generators of the cyclic group product.

Last edited: Nov 25, 2011
3. Nov 26, 2011

Hugheberdt

Thank you Dick for this somewhat different viewpoint!

I'd just like to check a few things to make sure I've got them right (I suspect they are correct, but just to check):

1.The elements 3 and 7 both have order 4, $\bullet$ modulo 20, so both would do as generators (but not 1 and 9, with orders 1 and 2) for a group isomorphic to C$_{4}$?

2.Since 3^0=1, 3^2=9, 3^3=7, a mapping of elements in H=(1,3,7,9, $\bullet$ mod20) to corresponding powers of generators in C$_{4}$ defines an isomorphism β: H -> C$_{4}$? E.i. β: β(1)=1, β(3)=x, β(9)=x^2, β(7)=x^3.

3. Is my initially proposed map β: β(1)=1, β(3)=x, β(7)=x^2, β(9)=x^3 invalid, since 7 is not 3^2 mod20?

4. Is this a correct isomorphism from U=(1,3,9,7,11,13,19,17,$\bullet$ modulo 20) to C$_{4}$$\times$C$_{2}$:
(u$\in$U, x$\in$C$_{4}$,y $\in$C$_{2}$)
β: β(3^n+10m)=(x^n,y^m) (0<n<3 , 0<m<1)?
If this is correct, can it be made more general or is some information redundant? (ommitting the restrictions on n and m or replacing 3 and 10 in the argument of β, I don't see how)
Of course I could map each element one by one, i.e. β(7)=(x^2,1) β(11)=(1,y), β(13)=(x,y), but in either case, how do I show that the groups are in fact isomorphic? By drawing the group table?

5. Is this a correct way of verifying β(ab)=β(a)β(b) for all a and b?
β: β((3^n+10m)(3^k+10l))=β(3^(n+k)+ (3^n)10l + (3^k)10m + 10l10m)=β(3^(n+k)+ (3^n)10l + (3^k)10m)= ???

Once again, thank you so much for your help!

4. Nov 26, 2011

Dick

It would probably read better if you use the additive notation for C4 and C2. So e.g. C4={0,1,2,3} with the operation being addition mod 4. If probably most direct to define an isomorphism f from C4xC2->G. Here's what I would do. The element (1,0) in C4xC2 has order 4. So we should map it into an element of G that has order 4. So pick f((1,0))=3. That means f((n,0))=3^n mod 20. Now we define f((0,1)) which has order 2. So we need that to be an element of order 2 in G that is NOT in the subgroup generated by 3. Let's pick 11. So f((n,m)) must be 3^n*11^m mod 20. In that form it's pretty easy to show it's an isomorphism.

Now let's take the isomorphism you discovered. g((n,m))=3^n+10*m. Can you figure out a way to show that f((n,m))=g((n,m)) for all n and m? You could also continue your proof to show it's an isomorphism directly by noticing that 3^n*10=10 mod 20. But I think the form f is a better way to express the isomorphism. It's more obvious that it is an isomorphism.

Last edited: Nov 26, 2011
5. Nov 26, 2011

Deveno

another way to do this:

say we have 2 abelian groups, G and AxB, where A and B are also abelian groups, and we want to show these two groups are isomorphic.

one way is to find two subgroups H and K of G isomorphic to A and B, respectively, such that H∩K = {e}. then if φ→A is one isomorphism, and ψ:K→B is the other, we have the map:

θ:G→AxB given by θ(hk) = (φ(h),ψ(k)).

since H∩K = {e}, G = HK, which justifies our definition of θ. because G is abelian, if hk = h'k', then hh'-1 = kk'-1, which implies both products are in H∩K = {e}, so: h'-1 = h-1, so h = h', and likewise with k and k'. so we can write every g in G uniquely as hk, for some h in H, and k in K.

comparing orders, we see θ is bijective, because both φ, ψ are bijective.

θ(gg') = θ((hk)(h'k')) = θ((hh')(kk')) (since G is abelian)

= (φ(hh'),ψ(kk')) = (φ(h)φ(h'),ψ(k)ψ(k')) = (φ(h),ψ(k))(φ(h'),ψ(k')) = θ(g)θ(g'),

so θ is an isomorphism.

perhaps you can see now, why it suffices to find an element x in U(20) of order 4, and another element y of order 2 in U(20) that is not in <x>. in this case, θ is:

θ(xkym) = (ak,bm), where a is a generator of C4, and b is a generator of C2.

Last edited: Nov 26, 2011
6. Dec 1, 2011

Hugheberdt

Thank you both!

You were both very helpful!

I feel like I've got a better grasp of it now.