Proving a partition is set of cosets Take a group $G$ with some partition $P$ such that for all $A,B\in P$, the product of the subsets $AB$ is contained entirely in some element $C$ of the partition.  Let $N\in P$ contain 1.  Prove $N$ is normal, and that $P$ is the set of cosets of $N$ in $G$. (Artin 2.10.3, trying to work through the whole book on my own)
It is pretty straightforward that $N$ is a normal subgroup.  Take another $A \in P$.  Because $N$ has the element $1$, then $A \subset AN$ and moreover $A = AN$ by the hypothesis. 
We would be done with the proof if we could show $aN = AN$ for an element $a \in A$.  Multiplying $A$ by any element in $N$ must preserve $A$, but why couldn't $A$ be a union of distinct cosets?
 A: Here i give another way: Define for any A,B in P, AxB=C, where C is the partition containing product set AB. Such a partition C is unique, since if not, you end up with more than one partitions containing AB. Try to show that (P,x) is group with N being its identity element.
 Now,consider the map f:G to P, defined by f(a)=A, where A is the partition in P containing a. This map is surjective  homomorphism with kernel N, as only elements in N are mapped to N.
So conclusions:
a) N is normal subgroup of G
b)Quotient group G/N=set of cosets of N is isomorphic to P and hence they are in bijective correspondence. 
A: Thanks for your help, here is my solution to tie up the thread.
Suppose we have a partition $P$ of group $G$, in which any product of two subsets of the partition $AB$ is contained in $C$, another subset in the partition.  Call $N$ the partition of $P$ that contains 1.  The product $NA$ will contain every element of $A$, since $N$ contains 1.  Therefore $NA = A$.  Likewise, $AN = A$, so $AN = NA$.  Also $NN = N$ necessarily because $NN$ is some subset in the partition and it contains 1, so it must be $N$.  Because the product $NN$ yields exactly the elements of $N$, it is closed under the group operation.  Suppose that $N$ didn't contain an inverse for some element $n \in N$.  Then $nN \subset NN$ contains $|N|$ distinct elements not including the identity, which is a contradiction because $|NN| = |N|$.  Therfore, $N$ is a subgroup.  Because $AN=NA$ for all $A\in P$, $N$ is a normal subgroup.  
Also, any coset $aN$ of $N$ is in $A$.  Because $A$ is closed under $N$, and the cosets of $N$ are precisely the sets preserved by $N$, $A$ must be the union of cosets.  
Suppose $A$ contains two disjoint cosets $aN$ and $a'N$.  Let $C\in P$ contain $a^{-1}$.  Therefore $AC$ contains 1, and $AC = N$.  However, this implies $a'a^{-1}=n$ for some $n\in N$, or $an^{-1} = a^{-1}$ which is a contradiction.
Therefore, each element of $P$ is a single coset of $N$.
