Showing a Projective Plane has 7 Points and 6 Lines

Question

I'm attempting to solve a little exercise that my maths lecturer has given me relating to projective planes. Basically, she has asked me to show that a projective plane contains a set of 7 points and 6 lines. I was wondering how this can be done, as using the axioms that define a projective plane, I can only achieve a Fano plane which contains 7 points and 7 lines as a circular line has to be included to satisfy the axioms.

The axioms in question are:

1. Any two distinct points lie on a unique line.
2. Any two distinct lines meet in a unique point.
3. There exist at least three non-collinear points.
4. There exist at least three points on every line.

Any tips or help will be greatly appreciated!

Answer 1

The exercise is wrong; any projective plane has as many lines as points. Let me try to give you the simplest proof from the axioms that $7$ points and $6$ lines is impossible.

There are $\binom72=21$ (unordered) pairs of points. Since each pair of points determines a line, if there are fewer then $7$ lines, then some line must be determined by more than $3$ pairs of points; i.e., some line must contain more than $3$ points. It will suffice to show that, if there are more than $3$ points on some line, then there are more than $7$ points all told.

Suppose that there are $4$ distinct points $Q_1, Q_2, Q_3, Q_4$ on the line $L.$ By Axiom 3, there is a point $P$ which is not on the line $L.$ Let $L_i$ be the line through $P$ and $Q_i.$ It's easy to see that the lines $L,L_1,L_2,L_3,L_4$ are all distinct. Each line $L_i$ must contain another point, call it $R_i,$ besides $P$ and $Q_i.$ Then the nine points $P,Q_1,Q_2,Q_3,Q_4,R_1,R_2,R_3,R_4$ are all distinct.

Answer 2

As commented above, I think the exercise was meant to show that any projective plane has at least 7 points and at least 6 lines. In fact, I think this is a good first exercise on projective planes to get some familiarity with the axioms.

By axiom 3, there exist three non-concurrent points $P_1$, $P_2$, $P_3$. As they are not concurrent, the lines $L_1 = \overline{P_1P_2}$, $L_2 = \overline{P_1P_3}$ and $L_3 = \overline{P_1P_2}$ (which exist by axiom 1) are distinct. By axiom 4, the three lines $L_i$ contain another point $Q_i$.

For $L_1$ that means $Q_1\notin\{P_2,P_3\}$. Moreover $Q_1 \neq P_1$, as otherwise the points $P_1$, $P_2$, $P_3$ would be collinear. In this way, none of the points $Q_1$ is contained in $\{P_1,P_2,P_3\}$.

Assume that $Q_1 = Q_2$. Then both lines $L_1$ and $L_2$ contain the points $Q_1 = Q_2$ and $P_3$. Now $L_1 = L_2$ by axiom 1, which is a contradiction. So $Q_1 \neq Q_2$ and in the same way $Q_1 \neq Q_3$ and $Q_2 \neq Q_3$. Therefore, the set $\{P_1,P_2,P_3,Q_1,Q_2,Q_3\} $is of size $6$.

Now for $i\in\{1,2,3\}$ define the lines $M_i = \overline{P_iQ_i}$. We have $M_1 \neq L_1$ as otherwise $P_1$, $P_2$ and $P_3$ would be collinear. Furthermore $M_1\neq L_2$, as otherwise $M_1 = L_2$ would contain $Q_1$ and $P_3$, so by axiom 1 $L_1 = \overline{Q_1P_3} = M_1$. Similarly, $M_1\neq L_3$. In the same way we see $M_i \notin\{L_1,L_2,L_3\}$ for all $i\in\{1,2,3\}$.

Also, $M_1 = M_2$ would contain $P_1$ and $P_2$, so by axiom 1 $M_1 = M_2 = L_3$ which whe have just seen to be impossible. Similarly, $M_1 \neq M_3$ and $M_2\neq M_3$. Altogether, $\{L_1,L_2, L_3, M_1, M_2, M_3\}$ is a set of $6$ lines.

By axiom 4, the line $M_1$ contains a point $R\notin\{P_1,Q_1\}$. The assumptoin $R = P_2$ or $R = Q_3$ would imply $M_1 = L_2$ by axiom 1 (as both lines contain $P_1$ and $R$), which is impossible. So $R\notin\{P_2, Q_3\}$ and similarly, $R\notin\{P_3, Q_2\}$. Hence $\{P_1,P_2,P_3,Q_1,Q_2,Q_3,R\}$ is a set of $7$ points.