Furnishing a contradiction in my proof involving Lagrange's Theorem

[jdinatale]

Homework Statement

The Attempt at a Solution

I'm trying to show that Case 1 implies that \tau \in H, and since \tau was an arbitrarily chosen 3-cycle, then H must contain all 3-cycles, thus contradicting that H has 6 elements.

 

[Deveno]

first of all, what is σ? it appears to be an arbitrary 3-cycle, but that isn't clear.

but going on that assumption, note that one of the elements of H is e.

therefore σ is in Hσ. similarly σσ is in Hσσ.

but since Hσ = Hσσ, σσ = τ_kσ, so σ = σσσσ = τ_kσσσ = τ_k.

thus σ is in H, which means <σ> = {σ,σσ,e} is in H.

but σ is an arbitrary 3-cycle, so...