Is an Odd Order Permutation Always an Even Permutation?

[Redhead711]

I have been working with the following question for quite awhile:

Show that a permutation with an odd order must be an even
permutation.

I have made some progress, but I am having trouble putting it altogether
to make my proof coherent.

This is what i have so far:

Let e= epsilon
Say BA^(2ka+1)= ae. Then BA^(2ka)=BA^(-1).
But BA^(2k)=(BA^ka)^2 is even.

I know that I am on the right track but I can't seem to put
it altogether. Can someone help me please. If I could
just have it explained Iam sure I will understand.

 

[HallsofIvy]

It would help if you told us what your notation, etc. meant.

1. How are you defining the "order" of a permutation?

2. Are A and B premutations? If so which is intended to be the "permutation with odd order?

3. What is k? what is a?

 

[mathwonk]

I suppose the order of a permutation is its order as an element of the group of permutations, i.e. that A^k = id for some odd number k>0, and for no smaller positive integer.

Then we claim A is "even". Recall that a product of an odd number of "odd" permutations is also "odd"...

does that help?

 

[Redhead711]

Yes you have helped me very much, I think I have a handle on the problem know than you both. :)