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Could I quickly spot the inverse of a permutation from its 2-cycle composition?

For example, given that $\rho=(1 \ 9)(1 \ 4)(1 \ 5)(1 \ 8)(2 \ 10)(2 \ 3)(2 \ 6)(2 \ 7)$, how to find its inverse from this 2-cycle decomposition?

Rescy_
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  • what happens if you multiply by the same transpositions in reverse order? – David Wheeler Sep 19 '15 at 12:34
  • @DavidWheeler, I see. So (1 8)(1 5)(1 4)(1 9)(2 7)(2 6)(2 3)(2 10) right? – Rescy_ Sep 19 '15 at 12:36
  • No, the reverse order would be $(2\ 7)(2\ 6)(2\ 3)(2\ 10)(1\ 8)(1\ 5)(1\ 4)(1\ 9)$, see Eclipse Sun's answer below... Ah, I see what you are doing, first you had disjoint cycles, which commute. You need to be careful about that, though. – David Wheeler Sep 19 '15 at 12:41

2 Answers2

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Yes, it is easy to write down the inverse of a permutation in this form. It is also not much harder to do this for any permutation given in cycle form.

There are two main points to note:

  1. If you have $c_1 \dots c_n $ then the inverse is given by $c_n^{-1} \dots c_1^{-1}$. Where $c_i^{-1}$ is the inverse of $c_i$. This is true in any group, not only for permutations.

  2. A $2$-cycle is its own inverse. So if each $c_i$ above is a two cycle then $c_n^{-1} \dots c_1^{-1}= c_n \dots c_1$.

If you want something similar without the condition that all cycles are $2$-cycles, it suffices to determine the inverse of a cycle can be obtained by reversing the order of the elements in the cycle, so if you have a cycle $c = (x_1 \ x_2 \dots x_k)$ then its inverse is $(x_k \dots x_2 \ x_1)$.

Note this is not a contradiction to the assertion that a $2$-cycle is its own inverse as $(x \ y)$ and $(y \ x)$ are the same permutation.

quid
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Hint: Use $(ab)^{-1}=b^{-1}a^{-1}$ and $(i\, j)=(i\, j)^{-1}$.

Where ab is the composite permutation $(a\circ b)(k)$, $k$ is an element of the set being permuted, and $(i\, j)$ represents the $2$-cycle of the elements $i$ and $j$.

Eclipse Sun
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  • You could at least say what is what. To solely rely on typographical details seems like a bad idea. – quid Sep 19 '15 at 12:58