# Groups :Ismorphism

1. Aug 11, 2011

### 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 $\in$U(t)

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

How to proceed beyond point ?

2. Aug 12, 2011

### Fredrik

Staff Emeritus
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?

3. Aug 13, 2011

### mehtamonica

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)

Last edited: Aug 13, 2011
4. Aug 14, 2011

### 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.

5. Aug 14, 2011

### 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.

6. Aug 14, 2011

### 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).

Last edited: Aug 14, 2011