PGL(2,3) 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 projective general linear group of degree two over field:F3 (the field of three elements) is isomorphic to symmetric group:S4.

Related facts

Similar facts

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

What the proof technique says for other projective general linear groups

The proof technique shows that there is an injective homomorphism from $PGL(2,q)$ (order $q^{3}-q$) into $S_{q+1}$ (order $(q+1)!$). In the cases $q=2$ ($PGL(2,2)\to S_{3}$) and $q=3$ ($PGL(2,3)\to S_{4}$) the orders on both sides are equal, hence the injective homomorphism is an isomorphism. For $q\geq 4$, the injective homomorphism is not an isomorphism.

Facts used

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

Proof

Step no.	Assertion/construction	Facts used	Previous steps used	Explanation
1	For any field $k$, $GL(2,k)$ has a natural action on the set $\mathbb {P} ^{1}(k)$ of all lines through the origin in $k^{2}$. The kernel of this action is precisely the scalar matrices.			[SHOW MORE] We know that $GL(2,k)$ acts on $k^{2}$ linearly, so it sends lines through the origin to lines through the origin. The scalar matrices send each line to itself. Conversely, suppose $g\in GL(2,k)$ fixes all lines. Let $v_{1},v_{2}$ be linearly independent vectors. Then, $g(v_{1})=\lambda _{1}v_{1}$ and $g(v_{2})=\lambda _{2}v_{2}$ for scalars $\lambda _{1},\lambda _{2}\in k^{\ast }$. If $\lambda _{1}\neq \lambda _{2}$, then $g$ does not send $v_{1}+v_{2}$ to a vector in the same line as $v_{1}+v_{2}$, hence does not preserve every straight line. This forces $\lambda _{1}=\lambda _{2}$, so $g$ acts as a scalar.
2	The action in Step (1) descends to a faithful group action of $PGL(2,k)$ on the set $\mathbb {P} ^{1}(k)$ of all lines through the origin in $k^{2}$, and hence an injective homomorphism from $PGL(2,k)$ to the symmetric group on the set $\mathbb {P} ^{1}(k)$.	Fact (1)	Step (1)	[SHOW MORE] follows because $PGL(2,k)$ is the quotient of $GL(2,k)$ by the scalar matrices, which are precisely the kernel of the action.
3	For $k$ a finite field of size $q$, $\mathbb {P} ^{1}(k)$ has size $q+1$.	Fact (2)		[SHOW MORE] The general formula for size of projective space of dimension $n$ over a field of size $q$ is $q^{n}+q^{n-1}+\dots +q+1$, or $(q^{n+1}-1)/(q-1)$. Plug $n=1$ to get $q+1$.
4	For $k$ a finite field of size $q$, $PGL(2,k)$, also denoted $PGL(2,q)$ has order $q^{3}-q$.	Fact (3)		[SHOW MORE] See Fact (3) for formulas. Explicitly, the order formula for $GL(2,k)$ is $(q^{2}-1)(q^{2}-q)$ (obtained by counting the number of possible vectors in the first row times the number of possible vectors in the second row conditional to the first vector). The scalar matrices form a subgroup of size $q-1$, so the quotient has size $(q^{2}-1)(q^{2}-q)/(q-1)=q(q^{2}-1)=q^{3}-q$.
5	For $k$ the field of size three, we have $\operatorname {Sym} (\mathbb {P} ^{1}(k))=S_{4}$ and its order is $24$ and $|PGL(2,k)|=|PGL(2,3)|=24$		Steps (3), (4)	[SHOW MORE] By Step (3), $\mathbb {P} ^{1}(k)$ has size $3+1=4$, so the symmetric group on it is $S_{4}$, and its order is $4!=24$. Also, by Step (4), $PGL(2,3)$ has order $3^{3}-3=24$.
6	For $k$ the field of size three, the injective homomorphism of Step (2) gives an isomorphism from $PGL(2,3)$ to $S_{4}$.		Steps (2), (5)	[SHOW MORE] Step (2) defines an injective homomorphism. Step (5) tells us that both the source and the target group have the same size. This forces the injective homomorphism to be surjective, and hence an isomorphism.

GAP implementation

The fact that the groups are isomorphic can be tested in any of these ways:

Command	Functions used	Output	Meaning	Memory usage
IsomorphismGroups(PGL(2,3),SymmetricGroup(4))	IsomorphismGroups, PGL, SymmetricGroup	[ (3,4), (1,2,4) ] -> [ (1,4), (1,2,3) ]	The output is an actual mapping of generating sets for the groups (as they happen to be stored in GAP) that induces an isomorphism. The fact that a mapping is output indicates that an isomorphism does exist. If the groups were not isomorphic, an output of fail would be returned.	22826
IdGroup(PGL(2,3)) = IdGroup(SymmetricGroup(4))	IdGroup, PGL, SymmetricGroup	true	The ID of a group is uniquely determined by the group's isomorphism class, and non-isomorphic groups always have different IDs. This test can thus be used to check whether the groups are isomorphic.	NA
StructureDescription(PGL(2,3))	StructureDescription, PGL	"S4"	The structure description of $PGL(2,3)$ is "S4" meaning that it is the symmetric group of degree four.	NA