Projective special linear group is simple

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:


 * $$n \ge 3$$.
 * $$k$$ has at least four elements.

Facts used

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

Related facts about special linear group and projective special linear group

 * Special linear group is perfect
 * Special linear group is quasisimple
 * Projective special linear group equals alternating group in only finitely many cases

Related facts about simplicity of linear groups

 * Projective symplectic group is simple
 * Projective special orthogonal group for bilinear form of positive Witt index is simple
 * projective special orthogonal group over reals is simple

Proof
The proof proceeds in the following steps:


 * 1) $$PSL_n(k)$$ 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.
 * 2) 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).
 * 3) $$PSL_n(k)$$ equals its own commutator subgroup when $$n \ge 3$$ or $$k$$ has at least four elements: This follows from facts (1) and (2).
 * 4) $$PSL_n(k)$$ is simple when $$n \ge 3$$ or $$k$$ has at least four elements: : This follows from the last two steps.