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anonymous

  • one year ago

Let S_n denote the group of all permutations of the set [n]={1,2,...,n}. A permutation sigma in S_n is called an involution if sigma^2 =1, where 1=(1) in S_n denotes the identity. Explicitly list all evolutions of S_1, S_2, S_3, S_4. Fore general n, show that a permutation 1 not equal to sigma in S_n is an involution if and only if sigma is a product of disjoint 2-cycles

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  1. anonymous
    • one year ago
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    How do you feel about part a?

  2. anonymous
    • one year ago
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    S_n are sufficiently simple for \(n\in\{1,\dots,4\}\) to manually list the involutions. an involution is essentially just a permutation that is its own inverse, i.e. it undoes itself when applied a second time

  3. anonymous
    • one year ago
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    proving that involutions in S_n are precisely products of disjoint 2-cycles is easy enough in the product of disjoint 2-cycles => involution direction, since they tell you that all 2-cycles are involutions. it is slightly less easy to show the other direction, namely that an arbitrary involution is a product of disjoint 2-cycles

  4. anonymous
    • one year ago
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    Perhaps you can begin by considering for each of the sets of permutations S_1, S_2, S_3, S_4. \[ [1] = \{1\} \\ [2] = \{1,2\} \\ [3] = \{1,2,3\} \\ [4] = \{1,2,3,4\}.\] Then \[S_1 = \{1\} \\ S_2 = \{1,(1,2)\} \\ \vdots\]

  5. anonymous
    • one year ago
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    I do not fully know how to complete part b

  6. anonymous
    • one year ago
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    Do you recognize that disjoint cycles commute with one another? Suppose then we have a permutation \(\sigma\) that can be factored as a product of disjoint 2-cycles \(\sigma_1\cdots\sigma_n\). It follows then that $$\sigma^2=\sigma_1\cdots\sigma_n\sigma_1\cdots\sigma_n=\sigma_1^2\dots\sigma_n^2=e^n=e$$since 2-cycles are obviously involutions, i.e. \(\sigma_1^2=\sigma_2^2=\dots=\sigma_n^2=e\), the identity; since they are disjoint, we were able to commute the \(\sigma_1,\dots,\sigma_n\) so that they square to the identity

  7. anonymous
    • one year ago
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    Okay, to find the converse do I assume a disjoint cycle and show that it is an involution?

  8. anonymous
    • one year ago
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    yes, exactly

  9. anonymous
    • one year ago
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    I am sorry but I am still confused with what you have written. Would you mind explaining it to me. Is that just one way or both ways of the proof.

  10. anonymous
    • one year ago
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    that's only one way

  11. anonymous
    • one year ago
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    Okay. SO that is saying that we have an involution sigma^2=(1) and leading to disjoint 2 cycles?

  12. anonymous
    • one year ago
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    maybe it would help if you did it the other way too. S o I can see the entire picture

  13. anonymous
    • one year ago
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    that is saying that a product of disjoint 2-cycles is an involution. to go the other way, suppose we have some permutation \(\sigma\). by cycle decomposition there is a more-or-less (i.e. up to commutativity) unique representation in products \(\sigma=\sigma_1\cdots\sigma_n\). if it's an involution we have \(\sigma^2=e\implies\sigma=\sigma^{-1}\) so $$\sigma_1\cdots\sigma_n=\sigma_n^{-1}\cdots\sigma_1^{-1}$$note the cycles are disjoint we can commute freely. do you see a way to deduce that \(\sigma_i=\sigma_i^{-1}\)

  14. anonymous
    • one year ago
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    That makes more sense now. Thanks a ton!

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