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Take the following pair of balanced partial triple systems $(R,P_1)$, $(R,P_2)$. Where $R=\{1,2,3,4,5,6\}$ and:

$P_1=\{\{1,3,5\},\{1,4,6\},\{2,4,5\},\{2,3,6\}\}$

$P_2=\{\{1,4,5\},\{1,3,6\},\{2,4,6\},\{2,3,5\}\}$

Suppose we have a steiner triple system of order $9$ denoted $(S,T)$ and we relabel $S$ so that $P_1\subset T$. If we replace $P_1$ in $T$ with $P_2$. Then we get a new steiner triple system of order $9$ denoted $(S,T')$. Furthermore, $|T\cap T'|=8$. However, in a theorem that I saw, there exists no pairs of $STS(9)'s$ with 8 triples in common. My question is what is wrong with this process?

My first thoughts were that $(S,T')$ might not be an $STS(9)$ when we swap $P_1$ for $P_2$ but it is one. Another thought I had is that it may be impossible to relabel $S$ so that $P_1\subset T$ but this doesn't make sense either since we can have $P_1\subset T$ without any relabelling. Any help is appreciated.

Ook
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1 Answers1

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Up to isomorphism there is only one Steiner triple system of order $9$: $$(1,2,3),(4,5,6),(7,8,9),(1,4,7),(2,5,8),(3,6,9),(1,5,9),(2,6,7),(3,4,8),(1,6,8),(2,4,9),(3,5,7)$$ This system has no subset of $4$ triples that contains only $6$ symbols, so the relabelling you propose is impossible. (The statement "we can have $P_1\subset T$ without any relabelling" is false.)

nickgard
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