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Different <math> \beta </math> give different <math>\sigma \beta </math>. Thus there are as many even permutations as there are odd ones. | Different <math> \beta </math> give different <math>\sigma \beta </math>. Thus there are as many even permutations as there are odd ones. | ||
| − | For each even permutation <math>\beta</math>, the permutation <math>\sigma \beta </math> in H is odd. | + | For each even permutation <math> \beta </math>, the permutation <math> \sigma \beta </math> in H is odd. |
Note: | Note: | ||
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<math>\sigma \beta </math> = odd | <math>\sigma \beta </math> = odd | ||
| − | Also, when <math>\sigma \beta </math> <math> \neq </math> <math> \beta \sigma </math> when <math> \sigma \neq \beta </math> | + | 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. |
Revision as of 15:18, 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.
