PGammaL(2,4) is isomorphic to S5

This article gives a proof/explanation of the equivalence of multiple definitions for the term symmetric group:S4
View a complete list of pages giving proofs of equivalence of definitions

Statement

The projective semilinear group of degree two over field:F4 (the field of four elements) is isomorphic to symmetric group:S5.

In symbols:

$P\Gamma L(2,4) \cong S_5$.

Related facts

Similar facts

• PGL(2,5) is isomorphic to S5
• PSL(2,5) is isomorphic to A5
• PGL(2,4) is isomorphic to A5
• PGL(2,3) is isomorphic to S4
• PSL(2,3) is isomorphic to A4

Facts used

1. Equivalence of definitions of size of projective space
2. Order formulas for linear groups of degree two

Proof

Step no.	Assertion/construction	Facts used	Previous steps used	Explanation
1	For any field $k$, there is a natural faithful group action of $P\Gamma L(2,k)$ on $\mathbb{P}^1(k)$, the set of lines through the origin in $k^2$, and hence an injective group homomorphism from $P\Gamma L(2,k)$ to the symmetric group on $\mathbb{P}^1(k)$.			Follows from definitions.
2	For $k$ the field of size $q$, $\mathbb{P}^1(k)$ has size $q + 1$ and thus the symmetric group on it has size $(q + 1)!$.	Fact (1)
3	For $k$ the field of size $q = p^r$ with $p$ prime, $P\Gamma L(2,k) = P\Gamma L(2,q)$ has size $r(q^3 - q)$.	Fact (2)
4	For $k$ the field of size $q = 4$, $P\Gamma L(2,k) =P\Gamma L(2,4)$ has order $2(4^3 - 3) = 2(60) = 120$ and $\operatorname{Sym}(\mathbb{P}^1(k))$ is the symmetric group of degree $4 + 1 = 5$ and has order $5! = 120$.		Steps (2), (3)	Plug in and evaluate.
5	For $k$ the field of size $q = 4$, the homomorphism of Step (1) gives an isomorphism from $P\Gamma L(2,4)$ to $S_5$.		Steps (1), (4)	By Steps (1) and (4), we get an injective homomorphism from $P\Gamma L(2,4)$ to $S_5$. Again by Step (4), both groups are finite and have the same order, hence the injective homomorphism must be an isomorphism.