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.