Projective special linear group is simple

This article gives the statement, and possibly proof, of a particular group or type of group (namely, Projective special linear group (?)) satisfying a particular group property (namely, Simple group (?)).

Statement

Let $k$ be a field and $n$ be a natural number greater than $1$ . Then, the projective special linear group $PSL_{n}(k)$ is a simple group provided one of these conditions holds:

$n\geq 3$ .
$k$ has at least four elements.

Facts used

Special linear group is perfect: Under the same conditions ( $n\geq 3$ or $k$ has at least four elements), $SL_{n}(k)$ is a perfect group: it equals its own commutator subgroup.
Perfectness is quotient-closed: The quotient of a perfect group by a normal subgroup is perfect.
Abelian normal subgroup of core-free maximal subgroup is contranormal implies commutator subgroup is monolith

Related facts

Proof

The proof proceeds in the following steps:

$PSL_{n}(k)$ satisfies the hypotheses for fact (3): Consider the natural action of $PSL_{n}(k)$ on the projective space $\mathbb {P} ^{n-1}(k)$ . This is a primitive group action, and the stabilizer of any point is thus a core-free maximal subgroup. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
The commutator subgroup of $PSL_{n}(k)$ is contained in every nontrivial normal subgroup of $PSL_{n}(k)$ : This follows from the previous step and fact (3).
$PSL_{n}(k)$ equals its own commutator subgroup when $n\geq 3$ or $k$ has at least four elements: This follows from facts (1) and (2).
$PSL_{n}(k)$ is simple when $n\geq 3$ or $k$ has at least four elements: : This follows from the last two steps.