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## Homework Statement

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.

## Homework Equations

## 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!