PSL(2,5) is isomorphic to A5

This article gives a proof/explanation of the equivalence of multiple definitions for the term alternating group:A5
View a complete list of pages giving proofs of equivalence of definitions

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. 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. 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).