Chauchy's Theorem is used to prove that finite groups with elements of order a power of

are of order

. However, in this case the elements are of order

. This is a simpler case, and we can use the formula,

That is to say, if we look at the problem in my proof each element has order

. As each element has order

it will generate

other elements of order

and so if we take a minimum generating set,

, the intersection of each pair is trivial,

and the formula can be applied. Thus,

and we have the contradiction needed to salvage my earlier proof.

I hope that makes sense...

It should be noted that that formula has an elementary proof, and so I believe that my proof will count as "elementary". I hope.