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A Steiner Triple System is a set $\mathcal{S}$ of $v \geq 3$ elements together with a set $\mathcal{B}$ of $3$-subsets (triples) of $\mathcal{S}$ such that every $2$-subset of $\mathcal{S}$ occurs in exactly one triple of $\mathcal{B}$. As an example, the following forms a Steiner Triple System of order $7$ on the set $\{ 1,2,3,4,5,6,7 \}$.

$\mathcal{S}_7= \{ \{ 1,2,4\},\{2,3,5\},\{3,4,6\},\{4,5,7\},\{5,6,1\},\{6,7,2\},\{7,1,3\}\}$

This does not satisfy the parallel line postulate (where lines are represented here by the elements of $\mathcal{S}_7$), which I am defining here as: "Given a line $l$ and a point $P$ not on $l$, there is at least one line through $P$ and parallel (disjoint) to $l$".

Consider the point $1$ and the "line" $\{2,3,5\}$. Every "line" containing $1$ has a non-empty intersection with $\{2,3,5\}$, and thus the parallel line postulate proves false.

However, for a Steiner Triple System of order $9$, the parallel line postulate holds true. My question is, does this hold true for all Steiner Triple Systems of order $\geq 9$? My gut says yes, but I'm struggling with proving it. Thanks in advance for the help.

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    Did you mean $9$ in the title? – 2017-02-14
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    I didn't want to use $\geq 9$ in the title so I used "greater than $7$", greater being strict in this sense. – 2017-02-14
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    If you're going to ask us a question about the "parallel line postulate", it might help to tell us what you mean by the "parallel line postulate". The answer might depend on that. Here are three guesses. (1) Given a line $\ell$ and a point $P$ not on $\ell,$ there is ***at most one*** line through $P$ and parallel to $\ell.$ (2) Given a line $\ell$ and a point $P$ not on $\ell,$ there is ***exactly one line*** through $P$ and parallel to $\ell.$ (3) Given a line $\ell$ and a point $P$ not on $\ell,$ there is ***at least one*** line through $P$ and parallel to $\ell.$ – 2017-02-22
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    Excellent point, I had forgotten I did not do that. Allow me to put it in the main post. – 2017-02-22

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Consider a Steiner triple system on $13$ points. Consider a line $\ell=\{a,b,c\}$ and a point $p\notin\ell.$ There are exactly $6$ lines containing the point $p.$ Exactly one of those lines contains $a,$ one of them contains $b,$ one of them contains $c,$ and the other three are "parallel" to $\ell.$ This might or might not be a counterexample to the parallel line postulate, depending on just what statement you mean by the "parallel line postulate."

In general, if the Steiner triple system has $n$ points, then there are $\frac{n-1}2$ lines through $p;$ three of those lines intersect the line $\ell,$ and the remaining $\frac{n-7}2$ are "parallel." So there are no parallels if $n=7,$ exactly one parallel if $n=9,$ more than one parallel in all other cases.

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    To clarify, the definition of "parallel line postulate" I am using here is "given a line $l$ and a point $P$ not on $l$, there is exactly one line through $P$ and parallel to $l$". – 2017-02-22
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    Oops, my mistake, I meant "at least one line". – 2017-02-22
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    Could you explain your answer more, please? How do you know there are $\frac{n-7}{2}$ parallel lines? – 2017-02-22
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    There are $\frac{n-1}2$ lines through $p.$ One of them contains $a,$ one of them contains $b,$ one of them contains $c,$ and the rest are parallel to $\{a,b,c\}.$ So the number of parallels is $\frac{n-1}2-3=\frac{n-7}2.$ – 2017-02-22
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    Alrighty! Thanks so much. :) – 2017-02-22