\chapter{Projective Structure on index $p$ normal subgroups}
Index $p$ normal subgroups\footnote{This is the conventional order of adjectives.}
of a group are a projective space over $\bZ/p$, namely
$$\Hom(G,\bZ/p)\setminus\set{0}/(\bZ/p)^*.$$
This is particularly interesting to me because finite group theory often feels
very un-geometric and very formal, but apparently has deep geometric structure,
as this example illustrates.
\section{Corollaries}
Given $2$ index $p$ normal subgroups, you get a projective line of $p+1$
subgroups, and we have notions of colinear, coplanar etc.
The number of index $p$ normal subgroups is the size of the projective space,
$1+p+\dots+p^n = \frac{p^{n+1}-1}{p-1}$.
In particular, the number of index $2$ normal subgroups must be $0,1,3,7,\dotsc$:
it cannot be 2, for instance.
\section{Proof}
Maps $G \to \bZ/p$ form a vector space (Hom inherits a structure from its target).
Given a non-zero map $f\from G \to \bZ/p$, its kernel is an index $p$ normal subgroup.
Its kernel is unchanged under multiplying $f$ by a unit in $\bZ^*$, so we get a map from
$\bP(\Hom(G,\bZ/p))$ to index $p$ normal subgroups.
Conversely, given an index $p$ normal subgroup $N \triangleleft G$,
the quotient is a group of order $p$, which is isomorphic to $\bZ/p$,
so we get a (non-zero) map $G \to G/N \cong \bZ/p$.
The isomorphism is not unique; it is only defined up to $\Aut(\bZ/p)=\bZ^*$
(isomorphisms are a torsor over automorphisms);
thus we get the inverse map from index $p$ normal subgroups to non-zero maps
$G \to \bZ/p$, up to $\bZ^*$.
These maps are obviously inverse: a class of maps gives a kernel,
a kernel gives a class of maps.
\section{Concretely for index 2}
Given two distinct index 2 subgroups\footnote{Which are necessarily normal.},
their symmetric difference is another index 2 subgroup; it's the third subgroup in the
line they define.
Each subgroup defines a map $G \to \bZ/2$, and their sum is a new map $G \to \bZ/2$.
Over $\bZ/2$, sum of characteristic functions of sets is the characteristic function of the
symmetric difference (or in this case the function is $1$ minus the characteristic function of the subgroup).
For $p\neq 2$ you take some combinations of $p$ intersections of the cosets of the subgroups
to get the other points on the projective line they define.
Indeed, for index 2 we can almost think of this as a group structure;
geometrically there are 3 points on a line, so given any 2 distinct points, you get the third.
(Hom is a group, and we're just removing the identity; $(\bZ/2)^*=1$
so we're not modding out).
It doesn't quite work because you can't add a point to itself
(you get zero, which we've taken out).