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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?

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I don't think you need to show $\,AN=aN\,$ for some $\,a\in A\,$, but rather that $\,A=xN\,$ , for some $\,x\in G\,$ – DonAntonio Sep 25 '12 at 4:07
Well, you're right, but $x$ must be in $A$. Also $aN \subset AN$ for any $a\in A$, which seems like a step in the right direction... – orlandpm Sep 25 '12 at 4:55
clearly, A is a union of cosets of N. Let A contain a and C contain a inverse ( sorry on my phone ). Then AC is in N. If a and b are in different cosets of N contained in A, then we have a contradiction because the product of b and a inverse is in N. – user641 Sep 25 '12 at 5:26

2 Answers 2

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.

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up vote 0 down vote accepted

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 $N$.

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You don't need the long argument about N being a subgroup. NN = N is enough. This follows from the one step subgroup test. – user641 Sep 25 '12 at 15:50

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