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.


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