PGammaL(2,4) is isomorphic to S5

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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

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.