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Homework Help: Finding a group isomorphism under modulo multiplication

  1. Nov 25, 2011 #1
    1. The problem statement, all variables and given/known data
    I am given the group (G,[itex]\bullet[/itex]) consisting of all elements that are invertible in the ring [itex]Z[/itex]/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[itex]_{4}[/itex] 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[itex]_{2}[/itex]: K to L , β[itex]_{2}[/itex](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)[itex]\bullet[/itex]b(mod 5))=x^a(mod 5)[itex]\bullet[/itex]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. jcsd
  3. Nov 25, 2011 #2


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    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
  4. Nov 26, 2011 #3
    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, [itex]\bullet[/itex] modulo 20, so both would do as generators (but not 1 and 9, with orders 1 and 2) for a group isomorphic to C[itex]_{4}[/itex]?

    2.Since 3^0=1, 3^2=9, 3^3=7, a mapping of elements in H=(1,3,7,9, [itex]\bullet[/itex] mod20) to corresponding powers of generators in C[itex]_{4}[/itex] defines an isomorphism β: H -> C[itex]_{4}[/itex]? 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,[itex]\bullet[/itex] modulo 20) to C[itex]_{4}[/itex][itex]\times[/itex]C[itex]_{2}[/itex]:
    (u[itex]\in[/itex]U, x[itex]\in[/itex]C[itex]_{4}[/itex],y [itex]\in[/itex]C[itex]_{2}[/itex])
    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!
  5. Nov 26, 2011 #4


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    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
  6. Nov 26, 2011 #5


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    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:H→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
  7. Dec 1, 2011 #6
    Thank you both!

    You were both very helpful!

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