Consider the set $\Omega_{n,k}:=\lbrace 1,\ldots,n\rbrace^k$. The set consisting of bijections of $\Omega_{n,k}$ into itself can be identified with the symmetric group $S_{n^k}$.

If $\sigma\in S_k$ and $\tau_1,\ldots,\tau_k\in S_n$, consider the bijection given by $f\colon \Omega_{n,k}\longrightarrow \Omega_{n,k}$ given by $(i_1,\ldots,i_k)\longmapsto\big(\tau_{\sigma(1)}(i_{\sigma(1)}),\ldots,\tau_{\sigma(k)}(i_{\sigma(k)})\big)$.

Question: Is there a bijection of the form above having sign $-1$?


If $k=1$, or if $n=k=2$, then the answer is yes. Otherwise, the answer is yes if $n$ is odd and no if $n$ is even.

Since the symmetric group is generated by transpositions, it is enough to look at the following two cases:

(i) $\tau_1$ is a transposition, and $\sigma$, $\tau_i$ for $i>1$ are all the identity; and

(ii) $\sigma$ is a transposition and all $\tau_i$ are the identity.

If $n$ is odd, then (i) and (ii) are both odd permutations, and when $n$ is even they are both even (except when $k=1$ or $n=k=2$).

  • $\begingroup$ For $n=k=2$, the map $(i,j)\longmapsto (j,i)$ is a transposition, so it is odd. Am I wrong? $\endgroup$ – Vincenzo Zaccaro Sep 17 '19 at 9:21
  • $\begingroup$ Yes you are right! But, provided that $m>1$, that is the only exception to what I wrote before. In general, for $n=2$ and $k>1$, the permutation I referred to as (i) has $2^{k-2}$ transpositions. $\endgroup$ – Derek Holt Sep 17 '19 at 13:33

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