Every Normal Subgroup Of An Contains A 3-Cycle

Suppose H is normal in An and H is not {e}, and n>4 Then H contains a three cycle.

Assume an element not equal to {e} from H as x = y1y2...ys Is a product of disjoint cycles yi and I may assume that the length of the cycles decrease so that, length(y1) >= length(y2) >= length(y3) >= .... >= length(ys)

I will write y1 = (a1,a2,a3,....,am)

CASE 1) if m>3

put W = (a1,a2,a3)

then Wx(W-1)(x-1) is in H. I.e.

(W)(y1y2...ys)(W-1)(ys-1)(ys-1-1)...(y2-1)(y1-1) = (W)(y1y2...ys)(W-1)(y1-1)(y2-1)...(ys-1)

(ie the yi commute)

Now, W contains only symbols from y1 so W commutes with y2 to ys

Wx(W-1)(x-1) = Wy1(W-1)(y1-1) = (a1,a2,a3)(a1,a2,a3,...,am)(a3,a2,a1)(am,...a2,a1) = (a1,a2,a4)(a3) = (a1,a2,a4), and therefore H contains a three cycle. //

CASE 2) length(y1) = length(y2) = 3

write y1 = (a1,a2,a3)

y2 = (a4,a5,a6)

put W = (a2,a3,a4)

Then Wx(W-1)(x-1) = (W)(y1)(y2)(W-1)(y1-1)(y2-1) as before

= (a2,a3,a4)(a1,a2,a3)(a4,a5,a6)(a4,a3,a2)(a3,a2,a1)(a6,a5,a4) = (a1,a4,a2,a3,a5) So H contains a cycle of length 5, so I reapply CASE 1 above, and therefore H contains a three cycle. //

CASE 3) length(y1) = m=3 and length(y2) = ... = length(ys) = 2

then x2 = (y12)(y22)...(ys2) = y12, which is also a three cycle in H. //

CASE 4) yi for all i are transpositions

y1 = (a1,a2), y2 = (a3,a4)

put W = (a2,a3,a4)

then Wx(W-1)(x-1) = (a2,a3,a4)(y1...ys)(a4,a3,a2)(y1...ys) = (a2,a3,a4)y1y2(a4,a3,a2)y1y2

(since y3...ys commute with (a4,a3,a2) as these only appear in y1 and y2 and the cycles are ALL disjoint.)

Wx(W-1)(x-1) = (a2,a3,a4)(a1,a2)(a3,a4)(a4,a3,a2)(a1,a2)(a3,a4) = (a1,a4)(a2,a3) which is in H

Now, I can take x = (a1,a4)(a2,a3) and since n>4 I can choose a5 not equal to a1, a2, a3 or a4 and I may then put W=(a1,a4,a5) and compute Wx(W-1)(x-1) in H

Wx(W-1)(x-1) = (a1,a4,a5)(a1,a4)(a2,a3)(a1,a5,a4)(a1,a4)(a2,a3) = (a1,a5,a4)(a2)(a3) = (a1,a5,a4) which is a three cycle in H. //

Proof so far complete. //

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