Module over an associative ring with unity and axioms of projective geometry According to Wikipedia(http://en.wikipedia.org/wiki/Projective_geometry#Axioms_of_projective_geometry), the axioms of projective geometry due to A. N. Whitehead are:
G1: Every line contains at least 3 points
G2: Every two points, A and B, lie on a unique line, AB.
G3: If lines AB and CD intersect, then so do lines AC and BD
 (where it is assumed that A and D are distinct from B and C).
Now let $V$ be a vector space over a field $K$.
We denote by $P(V)$ the set of one-dimensional subspaces of $V$.
If $V$ is finite dimensional, this is the usual definition of a projective space over $K$.
We say a two-dimensional subspace of $V$ a line of $P(V)$.
Then points and lines of $P(V)$ satisfy the above axioms(see
Whitehead's axioms of projective geometry and a vector space over a field)
Now let $M \neq 0$ be a left module over an associative ring $R$ with unity.
Suppose $M$ has a a composition series.
We also suppose that $M$ is not a finite set to avoid trivial cases. 
Then every submodule $N$ has a composition series.
We call its length the dimension of $N$ and denote it by dim $N$.
Let $P(M)$ be the set of one-dimensional submodules of $M$.
An element of $P(M)$ is called a point.
We say a two-dimensional submodule a line of $P(M)$.
Then it is clear that $P(M)$ satisfies Axiom G2.
Does $P(M)$ satisfy Axiom G1 or G3 or both?
If not, can you find counter-examples?
 A: It's clear that $G3$ always holds. If $A,B,C,D$ are four distinct simple submodules of $M$ such that $(A+B)\cap (C+D)\neq\{0\}$, then $A+B+C+D$ has length $3$. (You can see this by noting that $(A+B+C+D)/(A+B)\cong (C+D)/((A+B)\cap(C+D))$ and that the length of the right hand side is $1$ and then it must be that $len(A+B+C+D)-2=1$.
Then it is impossible for $(A+C)\cap (B+D)=\{0\}$, for then the sum $A+B+C+D$ would be direct, and we'd have a length $4$ module. So, the intersection is nontrivial.
$G1$ need not be true. Let $R=M=F_2\times F_3$, the product of fields of two and three elements, respectively. It has two isotypes of simple modules: $S_2=F_2\times\{0\}$ and $S_3=\{0\}\times F_3$ are representatives, and you can see that $M$ is just the line through those two points.
But $R$, as a cyclic group of order $6$, has exactly one subgroup of order $2$ and exactly one subgroup of order $3$. Therefore there can't be a third point on that line other than the two points we already spotted.
