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**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.