# Group Theory Problem

• October 8th 2011, 07:44 PM
Deveno
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.
• October 9th 2011, 01:01 PM
NonCommAlg
Re: Group Theory Problem
Quote:

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.

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

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

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

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

$g N \subseteq Q.$

let $Q_2 \in P$ be such that $N \subseteq g^{-1}Q \subseteq Q_2$. then $1 \in Q_2 \cap N$, because $1 \in g^{-1}Q \subseteq Q_2$, and so $Q_2=N$. hence $gN=Q$ and we are done.
• October 9th 2011, 01:29 PM
Deveno
Re: Group Theory Problem
G is not given to be finite, so $N = N^2$ is insufficient to establish that N is a subgroup.
• October 9th 2011, 02:19 PM
NonCommAlg
Re: Group Theory Problem
Quote:

Originally Posted by Deveno
G is not given to be finite, so $N = N^2$ 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!
• October 9th 2011, 02:44 PM
Deveno
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.
• October 22nd 2011, 12:16 PM
ymar
Re: Group Theory Problem
Quote:

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.