The set {1,2,...16} is to be split into two disjoint non-empty sets S and T in such a way that;

the product (mod 17) of any two elements of S lies in S;
the product (mod 17) of any two elements of T lies in S;
the product (mod 17) of any elements of S and any element of T lies in T.

Prove that the ONLY solution is

S={1,2,4,8,9,13,15,16} and T={3,5,6,7,10,11,12,14}.

I'd also like to know if there is a way to generalise this?

2. Originally Posted by Cairo
The set {1,2,...16} is to be split into two disjoint non-empty sets S and T in such a way that;

the product (mod 17) of any two elements of S lies in S;
the product (mod 17) of any two elements of T lies in S;
the product (mod 17) of any elements of S and any element of T lies in T.

Prove that the ONLY solution is

S={1,2,4,8,9,13,15,16} and T={3,5,6,7,10,11,12,14}.

I'd also like to know if there is a way to generalise this?
Generalization:

let $\displaystyle F$ be a finite field of odd order and $\displaystyle F^{\times}=F-\{0\}.$ we know that $\displaystyle F^{\times}$ is a cyclic group. there exist unique non-empty sets $\displaystyle S,T \subset \mathbb{F}^{\times}$ which satisfy the following conditions:

1) $\displaystyle S \cap T = \emptyset$ and $\displaystyle S \cup T=F^{\times},$

2) $\displaystyle a,b \in S \Longrightarrow ab \in S,$

3) $\displaystyle a,b \in T \Longrightarrow ab \in S,$

4) $\displaystyle a \in S, \ b\in T \Longrightarrow ab \in T.$

Proof: first note that 1) & 3) implies that $\displaystyle 1 \in S$ and thus by 4) we have $\displaystyle a^{-1} \in S,$ for all $\displaystyle a \in S.$ thus by 2), $\displaystyle S$ is a subgroup of $\displaystyle F^{\times}.$ now fix $\displaystyle a \in S, \ b \in T.$ see that $\displaystyle aS=S$ and $\displaystyle bS=T.$

thus $\displaystyle |S|=\frac{1}{2}|F^{\times}|,$ which proves the uniqueness of $\displaystyle S$ (and therefore the uniqueness of $\displaystyle T$), because we know that a cyclic group cannot have two subgroups of the same order. Q.E.D.

Remark: the above proof also gives the set $\displaystyle S$ (and therefore $\displaystyle T$): suppose $\displaystyle F^{\times}=<\alpha>.$ then $\displaystyle S=<\alpha^2>$ and $\displaystyle T=F^{\times}-S.$