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Also, when <math>\sigma \beta </math> <math> \neq </math> <math> \beta \sigma </math> when <math> \sigma \neq \beta </math>. In other words different <math> \beta </math> give different <math>\sigma \beta </math>. Thus there are at least as many odd permutations as there are even ones. | Also, when <math>\sigma \beta </math> <math> \neq </math> <math> \beta \sigma </math> when <math> \sigma \neq \beta </math>. In other words different <math> \beta </math> give different <math>\sigma \beta </math>. 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 <math> S_n </math>, form a subgroup <math> H \subset S_n </math> | ||
Revision as of 15:21, 9 September 2008
Question: 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 $
