I got this question in the exam: $T_{2n}$ is the subgroup of $S_{2n}$ that sends even numbers to even numbers and odd numbers to odd numbers, for example: $(2 4 6 8)(1 3 5)$ is a permutation in $T_{2n}$. what is the index of $T_{2n}$ in $S_{2n}$ ? Thanks for your help.
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Hint: $T_{2n} \cong S_{n} \times S_{n}$.
lhf
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The index is a famous number... – lhf Jan 28 '16 at 18:45
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OP here, sorry but can't comment yet (not enough reputation)
This is what I did, please correct my mistake: Order of $S_{2n}$ is $(2n)!$
Order of $T_{2n}:$ I figured there are $n!$ options for the evens and $n!$ for odds, but since these permutation are equal $(2\;\; 4\;\; 6\;\; 8) = (4\;\; 6\;\; 8\;\; 2)$ I figured, I should divide by $n$ both for evens and for odds. So I got: $$\frac {(2n)!}{(n-1)!\times (n-1)!}$$
userfault
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This answer is correct. If you are the OP (why a different username?) it's OK to answer your own question. Do learn the little $\TeX$ you need to format your questions and answers on this site. – Ethan Bolker Jan 28 '16 at 19:07
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My bad, I posted the question as a guest instead of my actual account. are you sure this is correct? most of my friends answered like lhf's answer below – userfault Jan 28 '16 at 19:14
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1My bad - @lhf is right. You don't divide by $n$ since you're counting the permutations, not how to represent them as cycles. – Ethan Bolker Jan 28 '16 at 19:20
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1You should try merging your accounts, then you will be able to comment on your question and edit it. – Daniel Fischer Jan 28 '16 at 20:33