A problem about exchanging apartments and permutation groups 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
 A: 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 $S_4$.
