GA(2,2) is isomorphic to S4

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 general affine group of degree two over field:F2 (the field of two elements) is isomorphic to symmetric group:S4.

Related facts

Similar facts

• GA(1,3) is isomorphic to S3
• PGL(2,2) is isomorphic to S3
• PGL(2,3) is isomorphic to S4
• PGL(2,5) is isomorphic to S5

Facts used

1. Order formulas for linear groups of degree two

Proof

Step no.	Assertion/construction	Facts used	Previous steps used	Explanation
1	For any field $k$ the group $GA(2,k)$ has a faithful group action on $k^2$ and hence has an injective homomorphism to the symmetric group on $k^2$.			By definition of $GA(n,k)$, it has a faithful group action on $k^n$.
2	For a field of size $q$, $GA(2,q)=GA(2,{\mathbb{F}}_q)$ has size $q^2(q^2-1)(q^2-q)$.	Fact (1)		[SHOW MORE] Follows directly from Fact (1). Explicitly: $GA(2,q)$ is a semidirect product of ${\mathbb{F}}_q^2$ and $GL(2,q)$. The former has order $q^2$ and the latter has order $(q^2-1)(q^2-q)$, so $GA(2,q)$ has order $q^2(q^2-1)(q^2-q)$.
3	For $k$ the field of size two, the symmetric group on $k^2$ is the symmetric group of degree four and its order is 24, and $GA(2,k)$ has order $2^2(2^2-1)(2^2-2)=24$.		Step (2)	[SHOW MORE] The order of $GA(2,k)$ can be computed by the formula in Step (2). The size of $k^2$ is $2^2=4$, so the symmetric group on it has order $4!=24$.
4	For $k$ the field of size two, the injective homomorphism of Step (2) gives an isomorphism from $GA(2,2)$ to $S_4$.		Steps (1), (3)	[SHOW MORE] By Step (1), there is an injective homomorphism from $GA(2,2)$ to $S_4$. By Step (3), both groups have the same finite order, so the injective homomorphism is an isomorphism.