230 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

Then for this permutation, i1 → i3, i2 → i2, i3 → i4, i4 → i1. The other numbers not in{i1, i2, i3, i4} are fixed, and in addition i2 is fixed which did not happen with α . Therefore,this new permutation moves only 3 numbers. Since it is assumed that m ≥ 4, this is acontradiction to α fixing the most points. It follows that

α = (i1, i2, i3)γ1 · · ·γ p (9.28)

or elseα = (i1, i2)γ1 · · ·γ p (9.29)

In the first case 9.28, say γ1 = (i4, i5, · · ·) . Multiply as follows α1 =

(i4, i2, i1)(i1, i2, i3)(i4, i5, · · ·)γ2 · · ·γ p (i1, i2, i4) ∈ B

Then form α1α−1 ∈ B given by

(i4, i2, i1)(i1, i2, i3)(i4, i5, · · ·)γ2 · · ·γ p (i1, i2, i4)γ−1p · · ·γ−1

1 (i3, i2, i1)

= (i4, i2, i1)(i1, i2, i3)(i4, i5, · · ·)(i1, i2, i4)(· · · , i5, i4)(i3, i2, i1)

Then i1 → i4, i2 → i3, i3 → i5, i4 → i2, i5 → i1 and other numbers are fixed. Thus α1α−1

moves 5 points. However, α moves more than 5 if γ i is not the identity for any i ≥ 2. Itfollows that

α = (i1, i2, i3)γ1

and γ1 can only be a transposition. However, this cannot happen because then the above α

would not even be in An. Therefore, γ1 = ι and so

α = (i1, i2, i3)

Thus in this case, B contains a 3 cycle.Now consider case 9.29. None of the γ i can be a cycle of length more than 4 since

the above argument would eliminate this possibility. If any has length 3 then the aboveargument implies that α equals this 3 cycle. It follows that each γ i must be a 2 cycle. Say

α = (i1, i2)(i3, i4)γ2 · · ·γ p

Thus it moves at least four numbers, greater than four if any of γ i for i≥ 2 is not the identity.As before, α1 ≡

(i4, i2, i1)(i1, i2)(i3, i4)γ2 · · ·γ p (i1, i2, i4)

= (i4, i2, i1)(i1, i2)(i3, i4)(i1, i2, i4)γ2 · · ·γ p ∈ B

Then α1α−1 =

(i4, i2, i1)(i1, i2)(i3, i4)(i1, i2, i4)γ2 · · ·γ pγ−1p · · ·γ−1

2 γ−11 (i3, i4)(i1, i2)

= (i4, i2, i1)(i1, i2)(i3, i4)(i1, i2, i4)(i3, i4)(i1, i2) ∈ B

Then i1→ i3, i2→ i4, i3→ i1, i4→ i3 so this moves exactly four numbers. Therefore, noneof the γ i is different than the identity for i≥ 2. It follows that

α = (i1, i2)(i3, i4) (9.30)