We can use the presentation ofSn+1to get a presentation ofAn+1, as follows. Let
αi = (1 2)(i i+ 1) =σ1σi, for 2≤i≤n.
Now letΓnbe the group with generatorsa2, . . . , anand relations
a32 =a21 = 1 for 3≤i≤n,
and
(aia−1j )
mij = 1 for all i6=j,
wheremij are as in (40).
Proposition 8.8 The mapai 7→αi extends to an isomorphismΓn
∼
−→An+1, so we have the presenta-
tion
An+1 ' ha2, . . . , an : a32 =a23 =· · ·=a2n= 1, (aia−1j )
mij = 1 ∀i6=j ∈[2, n]i.
Proof: We outline the proof, leaving some calculations to the reader. The elements bi = a−1i ∈ Γn
satisfy the same relations as the elementsai. Hence there is an automorphism ϑ : Γn → Γnsuch that
ϑ(ai) = a−1i .
LetΓ˜nbe the sethϑi ×Γnwith multiplication
(γ, ϑi)(γ0, ϑj) = (γ ·ϑi(γ), ϑi+j)23
In Γ˜ let s1 = (ϑ,1) and si = (ϑ, ai) for 2 ≤ i ≤ n. Then the {si} generate Γn and satisfy the
same relations as{σi} in Sn+1. Hence we have a surjection φ : Sn+1 → Γ˜n such that φ(σi) = si.
Reciprocally, the elementss1, α2, . . . , αn ∈ Sn+1 satisfy the same relations as {ϑ, a2, . . . , an} inΓ˜n.
Hence there is mapψ : ˜Γn→Sn+1which is the inverse ofφ. One checks thatψ(Γn) =An+1and this
completes the proof.
8.5.1 A presentation ofA5
The presentation of An in Prop. 8.8 has many relations, which can be inefficient. The following
well-known presentation ofA5 has fewer relations.
Proposition 8.9 We haveA5 ' ha, b|a5 =b3 = (ab)2 = 1}.
Proof: Let Γ = ha, b | a5 = b3 = (ab)2 = 1i.We first find elements α, β ∈ A5 obeying the same
relations as a, b. We may assume α = (1 2 3 4 5), and β = (i j k) is a 3-cycle such that αβ has order two. InA5, this meansαβ is a 221 cycle, so has a unique fixed-point. Making this fixed-point
1, we haveβ = (1 5 k). To have this be the only fixed-point of αβ we must have k = 3. Indeed,
(1 2 3 4 5)(1 5 3) = (2 3)(4 5). Henceβ = (1 5 3)works, and we have a homomorphismφ: Γ →A5
sendinga 7→ α,b 7→ β. The image of φcontains elements of orders2,3,5, hence is divisible by 30, butA5is simple, hence has no subgroups of order30, soφis surjective.
It remains to show that|Γ| ≤60. Consider the subgroupA =hai ≤Γ. Sincea5 = 1andφ(a) =α6=
1, it follows|A| = 5. It now suffices to show that|Γ/A| ≤12. This will be achieved if we can exhibit a set of12cosetsgAwhich is closed under multiplication byaandb.
The orbits ofAonΓ/Ahave size1or5, since5is prime. The groupB =hbiacts freely onΓ/A, since
3and5are relatively prime. We number the cosets according toa-orbits as follows.
1=A
2=bA, 3=abA, 4=a2bA, 5=a3bA, 6=a4bA,
7=a4ba4bA, 8=ba4bA=b·6, 9 =aba4bA, 10=a2ba4bA, 11=a3ba4bA, 12=ba3ba4bA=b·11,
We do not need to know if these cosets are distinct, since we only seek and upper bound on|Γ/A|. That1, . . . ,12are indeed distinct will result from the proof that they are closed under multiplication byaandb.
In the following diagram, a solid arrowi → j means thata·i= j, and imeansa·i = i. We will show that theb-action is given by the dashed arrows, wherei⇒jmeansa·i=j=b·i.
4 9 3 } } ? ? 10 1 ! ! 5 8 ^ ^ GG 12 c c 2 K S 11 ; ; ~ ~ 6 _ _ @ @ 7 W W O O
Since(ab)2 = 1andb−1 =b2, we have the relationaba=b2, hence
b·2=b2A=abaA=abA=3.
Sinceb·1=2andbhas order three, we have
b·3=b3·1=1,
so{1,2,3}is ab-orbit. Inverting the relationaba =b2, we getb =a−1b−1a−1. Using the knownb−
orbit{1,2,3}we compute
b·4=a−1b−1a−1·4=a−1b−1·3=a−1·2 =6.
Sinceb·6=8, it follows that{4,6,8}is a secondb-orbit. Then we have
b·5 =a−1b−1a−1·5=a−1·8=7,
b·9 =a−1b−1a−1·9=a−1·6=5,
so{5,7,9}is a thirdb-orbit. Finally,
b·10=a−1b−1a−1·10=a−1·7=11,
and sinceb·11=12, it follows that{10,11,12}is the fourth and finalb-orbit. It remains only to check thatafixes12:
a·12=aba3b·a4b1=aba3b·a5=aba·a·aba5=b−1ab−15=b−1a9=b−110=12.
This completes the proof that the cosets 1, . . . ,12 exhaust |Γ/A|. It follows that |Γ/A| = 12 and
Γ'A5as claimed.
Corollary 8.10 LetGbe a group containing nontrivial elementsx, y satisfying the relations
x5 =y3 = (xy)2 = 1.
Then the subgroup ofGgenerated by{x, y}is isomorphic toA5.
Proof: By Prop. 8.9 and the Mapping Property Prop. 8.4, there is a homomorphism φ : A5 → G
sendinga7→ x, b7→y. Andφis nontrivial sincex, y are nontrivial. SinceA5is simple,φis injective.
8.5.2 The exceptional isomorphismPSL2(9)'A6
Here is an illustration of the use of Cor.8.10. The fieldF9of nine elements can be constructed from the
fieldF3 =Z/3Zjust asCis constructed fromR, namelyF9 ={a+bi: a, b∈F3}with multiplication
determined by the rulei2 =−1. InSL2(9)the matrices
X = 1 +i i 1 1 , and Y = 1 1−i 0 1
satisfyX5 =−I,Y3 =I,(XY)2 =−I. Hence the imagesx, y ofX, Y inPSL
2(9)satisfy
x5 =y3 = (xy)2 = 1
and thereforex, ygenerate a subgroupH ≤PSL2(9)withH 'A5. Since|PSL2(9)|= 9(92−1)/2 =
360and|H|= 3·4·5 = 60, we have[PSL2(9) :H] = 6. The action ofPSL2(9)on the six cosets of
Hgives a homomorphism
ψ : PSL2(9)−→S6,
which is injective with image inA6, sincePSL2(9)is simple. Since|A6|= 3·4·5·6 = 360, it follows
thatψ gives an isomorphism
ψ : PSL2(9) ∼
−→A6,
which is one of the exceptional isomorphisms between linear and permutation groups.