Let $[n] = \{1,2,...,n\}$.
Suppose that $([n], T_1)$ and $([m], T_2)$ are $\mathrm{STS}$s on the sets $[n]$ and $[m]$ respectively. Let $T$ be the set of triples $\{(i, r),(j, s),(k, t)\}$ of elements in $[n] \times [m]$ that each satisfies one of the following conditions:
• $i = j = k$ and $\{r, s, t\} \in T_2$;
• $r = s = t$ and $\{i, j, k\} \in T_1$;
• $\{i, j, k\} \in T_1$ and $\{r, s, t\} \in T_2$.
Prove that $([n] × [m], T)$ is an $\mathrm{STS}$.
Using the example $([3], \{1,2,3\})$ and $([3], \{1,2,3\})$ it's pretty clear that the new $\mathrm{STS}$ is equivalent to $\mathrm{STS}(9)$. I've been stuck on this problem from from my textbook for a while. I'm lost on where to even begin proving this. Any hint would be appreciated.
Recall that a Steiner triple system $S(2,3,n)$ is collection of $3$ element subsets of $[n]$ (the blocks), such that every (distinct) pair (of elements of $[n]$) occurs in exactly one block.
The blocks of $S(2,3,[n] \times [m])$ are elements of the form $((i,r),(j,s),(k,t))$ and we need to check that each "pair of pairs" $((i,r),(j,s))$ occurs in exactly one block.
If $i=j$ we have the pair $((i,r),(i,s))$ and we know there is a block in $[m]$ that contains the pair that contains $(r,s)$ and so the first type of block in the definition of the product will contain this pair. And similarly if $r=s$ ...
So we need to chck the case $i \neq j$ and $r \neq s$, we know there is a block in $[n]$ that contains $(i,j)$ say $(i,j,k)$ and we know there is a block in $[m]$ that contains $(r,s)$ say $(r,s,t)$. By the third definition of the blocks of $[n] \times [m]$ ... $((i,r),(j,s),(k,t))$ is a block ... & we are done.