PSL(2,5) is isomorphic to A5

Statement
The projective special linear group of degree two over field:F5, which we denote $$PSL(2,5)$$, is isomorphic to alternating group:A5.

Facts used

 * 1) $$A_5$$ has order $$5!/2 = 60$$
 * 2) $$PSL(2,5)$$ has order $$(5^3 - 5)/2 = 60$$ by order formulas for linear groups of degree two
 * 3) uses::A5 is the unique simple non-abelian group of smallest order: This says that $$A_5$$ is the simple non-abelian group of smallest possible order and also that any simple non-abelian group of the same order as $$A_5$$ must be isomorphic to $$A_5$$.
 * 4) uses::Projective special linear groups are simple with some exceptions, but $$PSL(2,5)$$ is not among the exceptions, so $$PSL(2,5)$$ is simple.

Proof
The proof follows directly by combining Facts (1)-(4).