GA(2,2) is isomorphic to S4: Difference between revisions

Latest revision as of 03:35, 27 December 2021

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

Facts used

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 $G A (2, k)$ has a faithful group action on $k^{2}$ and hence has an injective homomorphism to the symmetric group on $k^{2}$ .			By definition of $G A (n, k)$ , it has a faithful group action on $k^{n}$ .
2	For a field of size $q$ , $G A (2, q) = G A (2, F_{q})$ has size $q^{2} (q^{2} - 1) (q^{2} - q)$ .	Fact (1)		[SHOW MORE] Follows directly from Fact (1). Explicitly: $G A (2, q)$ is a semidirect product of $F_{q}^{2}$ and $G L (2, q)$ . The former has order $q^{2}$ and the latter has order $(q^{2} - 1) (q^{2} - q)$ , so $G A (2, q)$ 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 $G A (2, k)$ has order $2^{2} (2^{2} - 1) (2^{2} - 2) = 24$ .		Step (2)	[SHOW MORE] The order of $G A (2, k)$ 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 $G A (2, 2)$ to $S_{4}$ .		Steps (1), (3)	[SHOW MORE] By Step (1), there is an injective homomorphism from $G A (2, 2)$ to $S_{4}$ . By Step (3), both groups have the same finite order, so the injective homomorphism is an isomorphism.

@@ Line 24: / Line 24: @@
 | 1 || For any field <math>k</math> the group <math>GA(2,k)</math> has a faithful group action on <math>k^2</math> and hence has an injective homomorphism to the symmetric group on <math>k^2</math>. || || || By definition of <math>GA(n,k)</math>, it has a faithful group action on <math>k^n</math>.
 |-
-| 2 || For a field of size <math>q</math>, <math>GA(2,q) = GA(2,\mathbb{F}_q)</math> has size <math>q^2(q^2 - 1)(q^2 - q)</math>. || Fact (1) || || <toggledisplay>Follows directly from Fact (1). Explicitly: <math>GA(2,q)</math> is a semidirect product of <math>\mathbb{F}_q^2</math> and <math>GL(2,q)</math>. The former has order <math>q^2</math> and the latter has order <math>(q^2 - 1)(q^2 - q)</math>, so <math>GA(2,q)</math> has order <math>q^2(q^2 - 1)(q^2 - q)</math>.
+| 2 || For a field of size <math>q</math>, <math>GA(2,q) = GA(2,\mathbb{F}_q)</math> has size <math>q^2(q^2 - 1)(q^2 - q)</math>. || Fact (1) || || <toggledisplay>Follows directly from Fact (1). Explicitly: <math>GA(2,q)</math> is a semidirect product of <math>\mathbb{F}_q^2</math> and <math>GL(2,q)</math>. The former has order <math>q^2</math> and the latter has order <math>(q^2 - 1)(q^2 - q)</math>, so <math>GA(2,q)</math> has order <math>q^2(q^2 - 1)(q^2 - q)</math>.</toggledisplay>
 |-
 | 3 || For <math>k</math> the field of size two, the symmetric group on <math>k^2</math> is the symmetric group of degree four and its order is 24, and <math>GA(2,k)</math> has order <math>2^2(2^2 - 1)(2^2 - 2) = 24</math>. || || Step (2) || <toggledisplay>The order of <math>GA(2,k)</math> can be computed by the formula in Step (2). The size of <math>k^2</math> is <math>2^2 = 4</math>, so the symmetric group on it has order <math>4! = 24</math>.</toggledisplay>