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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
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For each even permutation <math>\beta</math>, the permutation <math>\sigma \beta </math>  in H is odd.
 +
 
 +
Note:
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 +
<math>\sigma</math>  = even
 +
 
 +
<math>\beta</math> = odd
 +
 
 +
<math>\sigma \beta </math> = odd
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 +
Also, when <math>\sigma \beta </math>  <math> \neq </math>

Revision as of 15:15, 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 $

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett