Problem on Quotient Groups and First Isomorphism Theorem

Q: Let P be a partition of a group G with the property that for any pair of elements A, B of the partition, the product set AB is contained entirely within another element C of the partition. Let N be the element of P which contains 1. Prove that N is a normal subgroup of G and that P is the set of its cosets.

Sol: I am thinking of a surjective homomorphism f: G--> P with N as the kernel. Then N being the kernel is normal and G/N is isomorphic to P which implies P is the set of cosets of N in G.

My problem: P is a partiotion of G => there is an equivalence relation on G such that the elements of P are exactly the equivalence classes.

Now, if I define f: G--> P by f(a)= [a], where [a] represents the equivalence class of a,

then how can I prove [ab]=[a].[b] , i.e. f is a homomorphism?

Is my approach correct?

Please help.

1 Attachment(s)

Re: Problem on Quotient Groups and First Isomorphism Theorem

Hi,

Interesting little problem. I think a direct approach is probably the easiest; here's a solution:

Attachment 27873

Re: Problem on Quotient Groups and First Isomorphism Theorem

Quote:

Originally Posted by

**johng** Hi,

Interesting little problem. I think a direct approach is probably the easiest; here's a solution:

Attachment 27873

thanks...but is it possible to solve in my way of approach?

Re: Problem on Quotient Groups and First Isomorphism Theorem

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.)

Re: Problem on Quotient Groups and First Isomorphism Theorem

Quote:

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.)

thanks again for the reply...

so far we have f:G-->P a surjective homomorphism with kernel N. Now let for a, b in G, f(a)=f(b) i.e. a, b are both in the same partition [a]=[b]. Then, f(a(-1).b)=e [a(-1) is the inverse of a in G, and e is the identity element in P], since f is a homomorphism. So, a(-1).b is in N => b is in aN. Conversely, if b is in aN, then b=a.n for some n in N. Then, f(b)= f(a.n)= f(a). f(n) = f(a). e= f(a), i.e. b, a are in the same partition. Therefore, P is the set of cosets of N.

@johng do you think the above could be a valid proof?