1. ## Group Theory Problem

this isn't all that hard, but i decided to post it here, because it's interesting.

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 that contains the identity of G. Prove that N is a normal subgroup of G and that P is the set of its cosets.

2. ## Re: Group Theory Problem

Originally Posted by Deveno
this isn't all that hard, but i decided to post it here, because it's interesting.

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 that contains the identity of G. Prove that N is a normal subgroup of G and that P is the set of its cosets.

$\displaystyle (*)$ for any $\displaystyle A,B \in P,$ there exists $\displaystyle C \in P$ such that $\displaystyle AB \subseteq C.$

note that an obvious result of $\displaystyle (*)$ is that for any $\displaystyle Q_1 \in P$ and $\displaystyle g, h \in G$, there exists $\displaystyle Q_2 \in P$ such that $\displaystyle gQh \subseteq Q_2$.

now, for any $\displaystyle a \in N$ we have $\displaystyle a = a \cdot 1 \in N^2$ and so $\displaystyle N \subseteq N^2$. we also have $\displaystyle N^2 \subseteq Q$ for some $\displaystyle Q \in P$, by $\displaystyle (*)$. but then $\displaystyle N \subseteq Q$ and hence $\displaystyle N=Q$. therefore $\displaystyle N=N ^2$ and so $\displaystyle N$ is closed under multiplication. let $\displaystyle a \in N$. then $\displaystyle Na^{-1} \subseteq Q$ for some $\displaystyle Q \in P$ and since $\displaystyle 1 \in Na^{-1}$, we have $\displaystyle 1 \in N \cap Q$. thus $\displaystyle Q=N$ and so $\displaystyle Na^{-1} \subseteq N$. therefore $\displaystyle a^{-1} \in N$ and so $\displaystyle N$ is a subgroup. now let $\displaystyle g \in G$. then, by $\displaystyle (*)$, there exists $\displaystyle Q \in P$ such that $\displaystyle gNg^{-1} \subseteq Q$. but then $\displaystyle 1 \in Q \cap N$, because $\displaystyle 1 \in gNg^{-1} \subseteq Q$, and thus $\displaystyle Q=N$. so we have proved that $\displaystyle gNg^{-1} \subseteq N$, i.e. $\displaystyle N$ is normal.

finally, let $\displaystyle g \in G$ and choose $\displaystyle Q \in P$ such that $\displaystyle g \in Q$. then $\displaystyle gN \subseteq Q_1$ for some $\displaystyle Q_1 \in P$, by $\displaystyle (*)$. but then $\displaystyle g \in Q_1 \cap Q$, because $\displaystyle g \in Ng \subseteq Q_1$, and so $\displaystyle Q_1=Q$. hence

$\displaystyle g N \subseteq Q.$

let $\displaystyle Q_2 \in P$ be such that $\displaystyle N \subseteq g^{-1}Q \subseteq Q_2$. then $\displaystyle 1 \in Q_2 \cap N$, because $\displaystyle 1 \in g^{-1}Q \subseteq Q_2$, and so $\displaystyle Q_2=N$. hence $\displaystyle gN=Q$ and we are done.

3. ## Re: Group Theory Problem

G is not given to be finite, so $\displaystyle N = N^2$ is insufficient to establish that N is a subgroup.

4. ## Re: Group Theory Problem

Originally Posted by Deveno
G is not given to be finite, so $\displaystyle N = N^2$ is insufficient to establish that N is a subgroup.
ok, you're right but that can be easily fixed. see the solution again.
the problem is nice but the idea of the solution is very simple used over and over again!

5. ## Re: Group Theory Problem

what i found "interesting" about it, was not so much its difficulty level, but the idea that if a partition respects the group multiplication, not only must it be a co-refinement (is that even a word?) of the coset partition, but in fact they are equal.

i'm pretty sure that this is equivalent to saying that every congruence is in fact a coset partition.

and yeah, once you establish N is one element of the partition, g-translation pretty much wraps up the rest.

6. ## Re: Group Theory Problem

Originally Posted by Deveno
i'm pretty sure that this is equivalent to saying that every congruence is in fact a coset partition.
Indeed, if I'm not mistaken, the problem is a restatement of the well-known theorem stating the relation between congruences on groups and normal subgroups.

One of the two common definitions of a congruence on a semigroup is that it's an equivalence relation R such that, if (a R b) and (c R d), then (ac R bd). The hypothesis of the problem is clearly (I think but I'm often wrong) equivalent to the statement that P is a partition given by a congruence.