# Sign of some permutations

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.
(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 $$n=k=2$$).
• For $n=k=2$, the map $(i,j)\longmapsto (j,i)$ is a transposition, so it is odd. Am I wrong? – Vincenzo Zaccaro Sep 17 '19 at 9:21
• 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. – Derek Holt Sep 17 '19 at 13:33