Permutation Groups. Always Even or Always Odd.

I graduated nearly two and half years ago and I didn’t keep in touch with my math instructors, so I thought maybe some of you fine folk could help me out. I’m trying to keep my algebra skills sharp but I’m having a bit of difficulty reconciling these two theorems.

Now it seems to me that the lemma is saying that you cannot take the product of an odd number of 2-cycles, which would not only be a contradiction of the theorem, but is demonstrably false: (abcd) = (ad)(ac)(ab) is a product of an odd number of 2-cycles. The proofs of both the lemma and the theorem are in my textbook, and I can follow them both just fine (or at least I think I can :smack:). So what am I missing?

The lemma is saying that the identity permutation cannot be expressed as the product of an odd number of 2-cycles.

Now notice that the identity is an even permutation, since it can only be expressed as the product of an even number of 2-cycles:

For example:

as two 2-cycles: (12)(12)=e
as zero 2-cycles: e=e
(What you missed, of course, is that e is meant to represent the identity permutation.)

I’m not seeing where you’re getting that. What that lemma is saying is that if the decomposition is even, it’s always even, and if it’s odd, it’s always odd. The bits in parentheses mean that you actually have two different lemmas there, both of them true: You have the even version of the lemma, or you can replace both instances of the word “even” with the word “odd”, and have an odd version of the lemma which is also true.

But it’s not an e it’s an epsilon, and the textbook does use e as the identity. Perhaps it was a typo? The proofs also use epsilons. I’ll have to think about your answer but it seems right. Thanks for answering so fast!

The lemma says that r must be even, so doesn’t that imply an even number of β’s? For some reason the book has the lemma written before the theorem, maybe that’s what caused my confusion.

After re-reading the proof it seems that it was just a typo in the book. Epsilon should have been e, the identiy. Thanks again!