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...y_s Is a product of disjoint cycles y_i and I may assume that the length of the cycles decrease so that, length(y₁) >= length(y₂) >= length(y₃) >= .... >= length(y_s)

I will write y₁ = (a₁,a₂,a₃,....,a_m)

CASE 1) if m>3

put W = (a₁,a₂,a₃)

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

(W)(y₁y₂...y_s)(W^-1)(y_s^-1)(y_s-1^-1)...(y₂^-1)(y₁^-1) = (W)(y₁y₂...y_s)(W^-1)(y₁^-1)(y₂^-1)...(y_s^-1)

(ie the y_i commute)

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

Wx(W^-1)(x^-1) = Wy₁(W^-1)(y₁^-1) = (a₁,a₂,a₃)(a₁,a₂,a₃,...,a_m)(a₃,a₂,a₁)(a_m,...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(y_s) = 2

then x² = (y₁²)(y₂²)...(y_s²) = y₁², which is also a three cycle in H. //

CASE 4) y_i 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...y_s)(a₄,a₃,a₂)(y₁...y_s) = (a₂,a₃,a₄)y₁y₂(a₄,a₃,a₂)y₁y₂

(since y₃...y_s 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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