Originally Posted by

**johng** Hi,

Yes, I think so. But first if you want to define a homomorphism on G, the codomain of the function must be a group. That is, the partition P is a group. So you need to define the group operation on P. Let [a] denote the member of P containing group element a.

1. Define [a][b]=[ab]; you need to verify this is a well defined binary operation. That is if a1 is in [a] and b1 is in [b], then [ab]=[a1b1]. This is possible only because of the assumption on product sets and P.

2. All the other axioms for a group -- associativity, identity and inverses. This is all straight forward; in particular the identity is [1] where 1 is the identity element of G.

Note: for an arbitrary partition of P, the above "operation" is not well defined. Example: take P to be the right cosets of a subgroup H of G where H is not normal in G.

Now, you can define a homomorphism f on G by f(a)=[a]. By the definition of the binary operation on P, this is obviously a homomorphism. You can now verify that N is the kernel of f and so a normal subgroup of G. However you still must show the members of P are the cosets of N in G. I really don't see any other way to do this except as in my original posting. (Your assertion that G/N isomorphic to P implies this doesn't really follow.)