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This question is from Rotman's book An Introduction to the Theory of Groups.

How many $\alpha \in S_{n}$ are there with $\alpha^{2}=1.$ (Hint. (ij)=(ji) and (ij)(kl)=(kl)(ij))

My try: 1) There is identity permutation.

2) Transpositions: $\dfrac{n(n-1)}{2}$. ($2$ comes from the repetition in the transposition $(ij)$)

3) Product of two disjoint transpositions: $\dfrac{n(n-1)(n-2)(n-3)}{8}$ ($8$ comes from $2$ repetitions from the first transposition $2$ from the second and $2$ from the permutation of the two transpositions)

Even if my caslculations are correct I don't know how to get a closed formula for the summation.

I want to find the closed formula for the sumation and not counting the number of products of disjoint transpositions such that $\alpha^{2}=1$.

I am not a student, I'm just an old guy who likes to study Mathematics. I think I can live without knowing how to find the closed formula. Thanks for your hints.

  • Yes you are correct for those, but missing some more(not just two transpotitions). In general, $\alpha^2 = 1$ if and only if alpha is product of disjoint transpositions ($1$ can be regarded as an empty product) – Azlif Nov 09 '19 at 12:45

1 Answers1

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HINTS:

Try doing it in two steps.
1) Pick the elements that don't get fixed: $\binom{n}{2k}$
2) Find how many ways you can form pairs from step1: $\binom{2k}{2}$

$1+\binom{n}{2}+\binom{n}{4}\binom{4}{2} + \binom{n}{6}\binom{6}{2} + \cdots $

AgentS
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