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Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $\pi \in S_k$ such that for all $x_1, x_2 \cdots x_k \neq \mathbf{1} \in G$ we have that $x_1x_2x_3 \cdots x_k = x_{\pi(1)}x_{\pi(2)}x_{\pi(3)} \cdots x_{\pi(k)}$. Must $G$ be abelian?

This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(\mathbf{C})$ for counterexamples but also to no avail.

If this is false, is it possible that there's some conditions on the permutation $\pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.

MathematicsStudent1122
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1 Answers1

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Yes. By induction on $k\ge 2$; $k=2$ is clear.

Suppose $k\ge 3$. If $\pi(k)=k$ then it's clear by induction. Otherwise, write $j=\pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1\dots x_{k-1}=x_{\pi(1)}\dots x_{\pi(k-1)}$; write $\rho(i)=\pi(i)$ for $1\le i\le k-1$, $i\neq\pi^{-1}(k)$, $\rho(\pi^{-1}(k))=j$ ($\rho$ is the 1st return map on $\{1,\dots,k-1\}$); then $x_1\dots x_{k-1}=x_{\rho(1)}\dots x_{\rho(k-1)}$ for all $x_1,\dots,x_{k-1}$ not 1. Hence, if $\rho$ is not the identity, we are done.

Then we see that $\rho$ is the identity if and only if $\pi$ is the transposition $\tau_{j,k}$. Since the case when $\pi(1)=1$ is also clear, it remains to assume that $\pi=\tau_{1,k}$. So we have $x_1\dots x_k=x_kx_2\dots x_{k-1}x_1$ for all $x_i\neq 1$. If we find $x_2,\dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|\ge 3$ and then for $k\ge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.

YCor
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