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Sally's Journal
January 13th, 2005
04:54 pm


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That worked surprisingly well. I may push my luck and try it once more. Then I promise normal service will be resumed

"Show any subgroup of Sn which is not contained in An contains an equal number of odd and even elements."

This is question 3 on the sheet, so it's supposed to be easy, and has me flummuxed! I'm worried there's some trick like the only such subgroup being the whole group, but I can't see it...

(2 comments | Leave a comment)

[User Picture]
Date:January 13th, 2005 04:59 pm (UTC)
My notation is rusty, but am I right in thinking Sn is the group of all permutations of n elements, while An is the group of all even permutations of n elements?

If a subgroup G of Sn is not contained in An, this means it contains at least one odd element. Let E be such an element. Then consider the mapping f: G->G which maps x to Ex. This is a bijection (by group properties) and it maps every odd element to an even element and vice versa. Hence, there must be the same number of odd and even elements.
[User Picture]
Date:January 13th, 2005 05:03 pm (UTC)
... and Ex has to exist in the group because it's a subgroup and therefor closed....

I really am dim today. But I have some wonderful friends :-)
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