2
$\begingroup$

Suppose one has two $n$-tuples of complex numbers $(c_1,\dots,c_n)$ and $(z_1,\dots,z_n)$ such that all $c_i$, $z_i$ are nonzero, and $$ (c_1z_1,\dots,c_nz_n)=(z_{\sigma(1)},\dots,z_{\sigma(n)}) $$ and $$ (c_1^2z_1,\dots,c_n^2z_n)=(z_{\tau(\sigma(1))},\dots,z_{\tau(\sigma(n))}) $$ for some permutations $\sigma\ne\tau\ne1$.

Does this imply that all $c_i$'s are roots of unity?

Here's an example how this can happen: $\sigma=(123456)$, $\tau=(135)(246)$. Then one has \begin{align} z_1&\overset{c_1}{\longmapsto}z_2\overset{c_1}{\longmapsto}z_4\\ z_2&\overset{c_2}{\longmapsto}z_3\overset{c_2}{\longmapsto}z_5\\ z_3&\overset{c_3}{\longmapsto}z_4\overset{c_3}{\longmapsto}z_6\\ z_4&\overset{c_4}{\longmapsto}z_5\overset{c_4}{\longmapsto}z_1\\ z_5&\overset{c_5}{\longmapsto}z_6\overset{c_5}{\longmapsto}z_2\\ z_6&\overset{c_6}{\longmapsto}z_1\overset{c_6}{\longmapsto}z_3 \end{align} and we get the system for $c_i$'s: $$ c_1c_2=c_6,\quad c_2c_3=c_1,\quad c_3c_4=c_2,\quad c_4c_5=c_3,\quad c_5c_6=c_4,\quad c_6c_1=c_5 $$ and one can express $c_3,c_4,c_5,c_6$ as a product of powers of $c_1,c_2$ and then get $c_1^4=c_2^4=1$, so that all $c_i$'s are roots of unity. For example, the tuple of $c_i$'s can be $(i,i,1,i,-i,-1)$.

The question is: Is there a direct way to see that $c_i$'s are roots of unity, without drawing the graphs of permutations and solving corresponding system for the coefficients?

[Maybe if we denote $z=(z_1,\dots,z_n)$, $C=\operatorname{diag}(c_1,\dots,c_n)$, $P$ the permutation matrix for $\sigma$, and $Q$ the permutation matrix for $\tau\sigma$, then $$ C^{-1}Pz=z=C^{-2}Qz,\quad\text{ i.e. }\quad P^{-1}C^{-1}Qz=z $$ Since $P$ and $Q$ are unitary, this resembles the singular value decomposition. Any implications for the eigenvalues of $C$?]

$\endgroup$
0
$\begingroup$

Yes, this statement is true. It was proven in Theorem 4.2 in the article

Jack O. Button, Free by cyclic groups and linear groups with restricted unipotent elements. Groups Complex. Cryptol. 9 (2017), no. 2, 137–149.

This is also Theorem 6.2 in the arXiv paper:

J.O.Button, Minimal dimension faithful linear representations of common finitely presented groups. https://arxiv.org/abs/1610.03712

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.