PSL(2,4) is isomorphic to A5

Statement
Note that, due to the isomorphism between linear groups when degree power map is bijective, we know that over field:F4:

$$PGL(2,4) \cong PSL(2,4) \cong SL(2,4)$$

where $$PGL$$ denotes the projective general linear group of degree two, $$PSL$$ denotes the projective special linear group of degree two, and $$SL$$ denotes the special linear group of degree two.

The claim is that all these isomorphic groups are isomorphic to alternating group:A5.

Direct proof
The proof follows directly by combining (S1)-(S4) as follows: (S1) and (S2) establish that $$A_5$$ and $$PSL(2,4)$$ both have order 60. (S3) and (S4) establish that both are simple non-abelian, but $$A_5$$ is also the only one up to isomorphism for order 60 by (S3). This forces them to be isomorphic.