HW2, Chapter 5 problem 19, Discussion, MA453, Prof. Walther
Problem Statement:
Show that if H is a subgroup of $ S_n $, then either every member of H is an even permutation or exactly half of the members are even.
Answer:
Suppose H contains at least one odd permutation, say $ \sigma $. For each odd permutation $ \beta $, the permutation $ \sigma \beta $ is even.
Note:
$ \sigma $ = odd
$ \beta $ = odd
$ \sigma \beta $ = even
Different $ \beta $ give different $ \sigma \beta $. Thus there are as many even permutations as there are odd ones.
For each even permutation $ \beta $, the permutation $ \sigma \beta $ in H is odd.
Note:
$ \sigma $ = even
$ \beta $ = odd
$ \sigma \beta $ = odd
Also, when $ \sigma \beta \neq \beta \sigma $ when $ \sigma \neq \beta $. In other words different $ \beta $ give different $ \sigma \beta $. Thus there are at least as many odd permutations as there are even ones.
Conclusion: Since there are equal numbers of even and odd permutations, exactly half of the members are even.
By Thm. 5.6 in the text, "the set of even permutations in $ S_n $, form a subgroup $ H \subset S_n $. Therefore, if all members of H are even, we are done.
