Halmos Section 28: Parity
Point of post: In this post I will solve the, very few, problems in the twenty-eighth section of Halmos. This section is on the parity (sign) of a permutation.
Problem: How many elements of are there? (here denotes the Alternating Group)
Proof: Let be the set of all odd permutations in and define
We claim that this mapping is bijective. It’s clearly injective since . To see it’s surjective we note that if then and thus
so that . But, then . It follows that is bijective. Note though that since ( partition ) we may conclude that
and thus .
Problem: Give examples of even permutations with even order and even permutations with odd order; do the same for odd permutations.
Proof: First note that is an even permutation (as can be checked, I did it by computing and finding that its determinant is ) yet (or, now that I think about it, the identity map is even since but its order is ). Take , this is even and has even order.
Note that is odd but has even order. Take, . This is odd and has odd order.
Problem: Prove that generates .
Proof: We proceed by induction. Clearly this i true for . Now, suppose then that it’s true for . Then, note that if with , then is even and leaves fixed and thus by our induction hypothesis may be written as the product of -cycles, and thus may be written as the product of three cycles. The conclusion follows.
Remark: Note that in general that if that . To see this note that a quick computations shows that
from where it follows that can be written as as the product of row-changing matrices and thus
1. Halmos, Paul R. Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print
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