# Abstract Algebra: Homomorphism

1. Sep 23, 2009

### dav3

Hey all,
Ive been working at this "proof" for several hours now, have put it away several times thinking that maybe Ill get it if I leave it alone for a bit...has not worked =] It has 2 parts, I think I have proven the first part, but the second one really just stumps me =|

1. Show that a residue class [a] module mn is a subset of the residue class [a] modulo m. This defines a homomorphism Z/mn $$\rightarrow$$ Z/m, where Z is the set of integers. A similar construction gives a homomorphism from Z/mn to Z/n.

Now, use the above maps to construct a homomorphism Z/mn $$\rightarrow$$ Z/m x Z/n coordinatewise, i.e., from f:R $$\rightarrow$$ S and g:R$$\rightarrow$$ T, each x $$\in$$ R determines (f(x),g(x)) $$\in$$ S x T. This must shown to be a homomorphism if f and g are.

Okay, I believe I have a "proof" of the first part. Here it goes:

A residue class [a] mod mn contains all integers b such that mn | (a-b). Now, by definition, m | mn. Since m|mn and mn|(a-b), we see that m|(a-b) (if a|b and b|c then a|c). Hence, all integers that are members of [a] mod mn are members of [a] mod m, so [a] mod mn is a subset of [a] mod m.

I believe the same method is used to prove that [a] mod mn is a subset of [a] mod n. QED (i think =))

As for the second part, I don't even know where to begin. Any input/help would be greatly appreciated!

2. Sep 24, 2009

### aPhilosopher

Well your problem with the second part might be that you haven't finished the first part.

What are the maps that you're supposed to get out of the first part? The notation $$[a]_{m}$$ = residue class of a mod m is useful for defining the map. It's not clearly stated that you need to show that this is a homomorphism in the first part but proving that it is might be helpful anyway unless you already see it.