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The abstract of this paper says:

"It is well-known that any permutation can be written as a product of two involutions."

I was looking for any web resource that can provide an affirmation and (hopefully easy) proof of this statement -- can anyone please help?

And if any permutation can indeed be written as a product of two involutions, are the following guesses correct?

  1. If $P$ is a permutation and $X$ & $Y$ are involutions, and $P = XY$, then $P^{-1} = YX$
  2. If $X$ & $Y$ are distinct involutions such that neither is the identity permutation $I$, then the permutation $XY$ is not an involution.
  3. The only ways to express any involution $X$ as a product of two involutions is $X = XI$ & $X = IX$ (given that $I$ itself is an involution)

Thanks ...

Avijit
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    A general factoid at work in Michael's excellent answer is the following. If $\alpha=(a_1a_2a_3\ldots a_m)$ is an $m$-cycle and $\beta=(b_1b_2b_3\ldots b_n)$ is an $n$-cycle DISJOINT FROM $\alpha$, then multiplying the product $\alpha\beta$ by the transposition $(a_1b_1)$ "fuses" them to an $(m+n)$-cycle: $$ \alpha\beta(a_1b_1)=(a_1b_2b_3\ldots b_mb_1a_2a_3\ldots a_n). $$ He is using this repeatedly to form a cycle of a prescribed length as a product of two sets of disjoint 2-cycles. At least that's how I read it :-) – Jyrki Lahtonen Sep 19 '13 at 05:29
  • @JyrkiLahtonen Is there a name for this factoid, or any proof of it. Seems proof should be only by induction. – jiten Dec 01 '22 at 03:03
  • @JyrkiLahtonen Please tell why swapped the subscripts ($m,n$) of the two strings: $\alpha, \beta.$ Am not clear, if the swap is correct even. – jiten Dec 01 '22 at 03:38
  • @JyrkiLahtonen Let, $\alpha=(12345)$ \begin{pmatrix} 1&2&3&4&5\ 2&3&4&5&1\ \end{pmatrix}

    Let, $\beta=(6789)$ \begin{pmatrix} 6&7&8&9\ 7&8&9&6\ \end{pmatrix}

    Then, your comment means: $\alpha\beta(16)= (1789623)=$ $$1\to7, 7\to8, 8\to9, 9\to6\to 1, 6\to2, 2 \to3.$$

    But, have: $(12345)(6789)(16)=$ $$1\to 6\to7, 6\to1\to 2, 7\to8, 8\to9, 9 \to6, 2\to3, 3\to4, 4\to5, 5\to1.$$ So, it should be: $\alpha\beta(16)= (178962345).$

     – jiten Dec 01 '22 at 04:09
  • @jiten There was a typo with the subscripts. We get an $(m+n)$-cycle. Comments cannot be edited, so such typos linger. In your example case we get a 9-cycle. – Jyrki Lahtonen Dec 01 '22 at 05:03
  • @jiten I give a proof of this factoid as an exercise. It is not too difficult. Try it! – Jyrki Lahtonen Dec 01 '22 at 05:05

1 Answers1

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For your three numbered statements, $(1)$ is true (easy proof), but $(2)$ and $(3)$ are false. A counter example for $(2)$ is obtained by taking $X = (1,2)$ and $Y = (3,4)$. Then $XY = (1,2)(3,4)$ is an involution as well. A counterexample for $(3)$ is obtained from this example as well; the involution $X = (1,2)(3,4)$ can be factored as $YZ$ where $Y = (1,2)$ and $Z = (3,4)$.

As for the statement in question, here's a quick proof sketch:

(1) By using the disjoint cycle decomposition, you can reduce to proving that the cycle $(1,2,3,\dots,n)$ can be written as a product of two involutions in $S_n$.

(2) To handle that case, draw $n$ vertices in the plane (labelled $1,2,\dots,n$) and connect the $n$ vertices by drawing $n-1$ edges. This will make a unique (up to choice of direction to travel) path in your graph. Label the edges $1,2,\dots,n-1$ in the order of the path. For each edge, put the two vertices connected by that edge into a two-cycle. Then form $\pi_1$, the product of the two-cycles formed in this way from odd-numbered edges, and $\pi_2$, the product of the two-cycles formed in this way from even-numbered edges. Then the product $\pi_2 \pi_1$ is an $n$-cycle $\tau$. This needs to be checked; in fact, if you number the vertices in the order of the path, then $\tau = (1,3,5,\dots, 6,4,2)$. Conjugate the relation $\tau = \pi_2 \pi_1$ to get that $(1,2,\dots,n)$ is a product of two involutions.

Michael Joyce
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  • For finite reflection groups, we know quite a lot about involutions... For instance it is comparatively easy to find their conjugacy classes. Is there a similar theorem there? (about the number of involutions needed to factorize an arbitrary element of the reflection group...) – Theo Douvropoulos Mar 31 '15 at 05:17