Group Isomorphism: Proving f: Us(st)->U(t) is Onto

[mehtamonica]

To prove that : f : U_{s} (st) \rightarrow U(t) is an onto map.

Note that

Us(st)= {x \in U(st): x= 1 (mod s)}

Let x \inU(t)

then (x, t)=1 and 1<x< t

How to proceed beyond point ?

 

[Fredrik]

I don't understand what you're saying. How do you define the sets (groups?) U(t) and U(st) and? Is s an integer? How is the function f defined?

 

[mehtamonica]

I don't understand what you're saying. How do you define the sets (groups?) U(t) and U(st) and? Is s an integer? How is the function f defined?

1) U ( n) is a multiplication group modulo n, for any integer n.

( U(n) is group contains all the non zero units of Zn, that is , all the integers belonging to Zn that are relatively prime to n.

2) s and t are integers such that (s, t) =1.( rel prime)

3) f (x) = x mod s, where x belongs to Us(st)

 

[HallsofIvy]

But why are you limiting yourself like that? For any group, G, the function f_s(t)= st (or in "additive" notation, f_s(t)= s+ t) is an isomorphism- and so both one-to-one and onto.

To prove it is onto, suppose y is a given member of the group. You merely need to show that there exist x such that sx= y- and that is easy.

 

[Stephen Tashi]

This might be a question about a mapping between the multiplicative group of integers mod st to the multiplicative group of integers mod t rather than a question about mapping the multiplicative of integers into itself. In fact, from the definition of U_s in the original post, it might be a question about mapping subsets of the group of integers mod st onto the group of integers mod t.

 

[daveyp225]

Do you know U(st) is isomorphic to U(s)xU(t) though the map x->(x mod s, x mod t)?

Then Us(st) is a subgroup of U(st). Images of subgroups are subgroups, and in this case the image is {1} x U(t) which is isomorphic to U(t).