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| − | + | =[[HW2_MA453Fall2008walther|HW2]], Chapter 5 problem 19, Discussion, [[MA453]], [[user:walther|Prof. Walther]]= | |
| − | Answer: | + | ==Problem Statement:== |
| − | Suppose H contains at least one odd permutation, say <math>\sigma</math>. For each odd permutation <math>\beta</math>, the permutation <math>\sigma \beta</math> | + | |
| + | Show that if H is a subgroup of <math>S_n</math>, 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 <math>\sigma</math>. For each odd permutation <math>\beta</math>, the permutation <math>\sigma \beta</math> is even. | ||
| + | |||
| + | Note: | ||
| + | |||
| + | <math>\sigma</math> = odd | ||
| + | |||
| + | <math>\beta</math> = odd | ||
| + | |||
| + | <math>\sigma \beta </math> = even | ||
| + | |||
| + | 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. | ||
| + | |||
| + | Note: | ||
| + | |||
| + | <math>\sigma</math> = even | ||
| + | |||
| + | <math>\beta</math> = odd | ||
| + | |||
| + | <math>\sigma \beta </math> = odd | ||
| + | |||
| + | Also, when <math>\sigma \beta \neq \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>. | ||
| + | Therefore, if all members of H are even, we are done. | ||
| + | ---- | ||
| + | [[HW2_MA453Fall2008walther|Back to HW2]] | ||
| + | |||
| + | [[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008 Prof. Walther]] | ||
Latest revision as of 15:37, 22 October 2010
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.
