Yahoo Answers: Answers and Comments for Let G be Sn, the symmetric group of order n, acting as permutations on the set {1,2,...n}. Let H =? [Mathematics]
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From Dan
enUS
Wed, 03 Apr 2013 22:43:00 +0000
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Yahoo Answers: Answers and Comments for Let G be Sn, the symmetric group of order n, acting as permutations on the set {1,2,...n}. Let H =? [Mathematics]
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From wahlers: that is an similar element because the nonexi...
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Tue, 25 Oct 2016 15:42:35 +0000
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From ?: (i) This is trivial, but writing it down is fi...
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https://answers.yahoo.com/question/index?qid=20130403224300AA3JDXR
Wed, 03 Apr 2013 23:06:22 +0000
(i) This is trivial, but writing it down is fiddly. It would look better with bold chracters, but I'll make do with apostrophes. Please read X' as "The symbol X in boldface".
I'll let Sn1 act as permutations on the set { 1', 2', ..., (n1)' } and for each s ∈ H define the function s' : { 1', 2', ..., (n1)' } as follows:
s'(k') = s(k).
You can prove that s' is a bijection (permutation) as follows:
s'(k') = s'(l') <=> s(k) = s(l) <=> k = l, so s' is injective.
For every l' ∈ { 1', ..., (n1)' }, we have l ∈ { 1, ..., n1 }, and so there is some k ∈ { 1, ..., n1 } with s(k) = l, and so s'(k') = l'. This means that s' is surjective.
Now to show that the map s > s' is an isomorphism: Let s,t ∈ H, then
(s' o t')(k') = s'(t'(k')) = s(t(k)) = (s o t)(k) = s(k) t(k) = s'(k') t'(k').
(ii) For k ∈ { 1, ..., n1 } set ak = (k n), and let an be the neutral permutation in G.
(iii) The cosets are given by H(ak) = { s ∈ G : s(n) = k }. Note that
H = H(an) = { s ∈ G : s(n) = n } as you wanted.