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Every pair of inhabitants of city Z may exchange their apartments once a day (X moves to apartment Y, Y moves to apartment X). Is it possible to make any complicated apartment exchange in only two days?

Don’t really know how to start with this group theory riddle, it seems like it relates to permutation groups, but I’m unsure. Any advice is appreciated

kabenyuk
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Sigurd Solskjær
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  • Let $|Z|$ be the number of people in $Z$. I would start by considering $S_{|Z|}$. Think about its generators. – Shaun Nov 28 '22 at 19:46
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    Assuming that inhabitants may or may not exchange apartments, this is equivalent to saying that any element of $S_{n}$ is a product of two involutions. See for instance this MSE post or this arXiv manuscript. – Andreas Caranti Nov 29 '22 at 12:40
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    Let me add that this is really a problem about dihedral groups. Considering a permutation as the product of disjoint cycles (see the post quoted in the previous comment), you only need to do the case when the permutation is $\sigma=(123…n)$. If $\langle \sigma, \tau \rangle$ is the dihedral group of order $2n$, where $\tau$ is an involution, just note that $\sigma \tau$ is a reflection, and thus another involution, so that $\sigma = (\sigma \tau) \tau$ is the product of two involutions. For instance we may take $\tau=(2, n)(3,n−1) \dots$ as $\tau \sigma \tau = \sigma^{-1}$. – Andreas Caranti Nov 29 '22 at 13:29
  • Addendum. And then $\sigma \tau = (1, n) (2, n-1) \dots$. – Andreas Caranti Nov 29 '22 at 13:35

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As I understand it, none of the residents of $Z$ can refuse to be relocated. If this is true, then in the language of permutations this problem is equivalent to the following.

Can any permutation of $S_n$ ($n=|Z|\geq4$) be represented as a product of two permutations, each of which has the form $$ (z_1z_2)\ldots(z_{n-1}z_n).\tag1 $$

(Since no one can refuse the relocation, $n$ is even.)

The answer to this question is negative. First, every product of the form $\sigma\tau$, where both $\sigma$ and $\tau$ have the form $(1)$, is necessarily an even permutation.

Second, not every even permutation is represented in this form. This can be proved like this.

Let $k=n/2$ for brevity. The number of permutations of the form $(1)$ is $$ \frac{1}{k!}\cdot\binom{n}{2}\binom{n-2}{2}\ldots\binom{4}{2}=\frac{n!}{2^kk!}. $$ By induction on $n$ it is proved that $$ \left(\frac{n!}{2^kk!}\right)^2<\frac{n!}{2} $$ for any even $n\geq4$.

Thus the number of products $\sigma\tau$ of permutations of the form $(1)$ is less than $n!/2$.

In the simplest case of $n=4$ there are only $3$ of permutations of the form $(1)$. Together with the identity they form a subgroup of order $4$ in $S_4$.

kabenyuk
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