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Let $S_n$ be the symmetric group on $n$ elements $\{ 1,2,\dotsc,n \}$ and $G$ be a subgroup of $S_n$ of order $n$. Denote the elements in $G$ by $\{ \sigma_1,\dotsc,\sigma_n \}$. Let the matrix $A=(\sigma_i(j))\in M_{n\times n}(\mathbb Q)$. In general, $A$ is not invertible as pointed out by Dave.

If $G$ is a transitive subgroup of order $n$, is the matrix $A$ invertible?

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I presume you mean invertible in $M_{n\times n}(\mathbb{Q})$. Here's an example. $$\left(\begin{matrix} 1&2&3&4&5&6\\ 2&1&3&4&5&6\\ 1&3&2&4&5&6\\ 3&1&2&4&5&6\\ 2&3&1&4&5&6\\ 3&2&1&4&5&6 \end{matrix}\right)$$ The last three columns are linearly dependent.

A slightly more interesting question is whether it's true for a transitive (and hence regular) subgroup of order $n$.

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    $\begingroup$ So the group $G$ in your example is $S_{3}$ acting naturally on $\{1,2,3\}$ and fixing $\{4,5,6\}$ elementwise? $\endgroup$
    – Geoff Robinson
    Commented Nov 14 at 16:22
  • $\begingroup$ @Dave, thanks a lot for your example! I revised the matrix to be valued in $\mathbb Q$. $\endgroup$
    – lin
    Commented Nov 14 at 17:47
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The answer is no. Consider the subgroup generated by (1243). The matrix you have will be cyclic. See Determinant of cyclic matrix, proof without eigenvectors for a discussion of the determinant.

It is left as an exercise that when $f(x)=1+2x+4x^2+3x^3$, we have $f(-1)=0$, and $-1$ is a 4-th root of unity.

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    $\begingroup$ There are plenty of transitive permutation groups on $n$ points not having an $n$-cycle. For example, the Klein group $\langle (1,2)(3,4), (1,3)(2,4)\rangle$. $\endgroup$
    – Mark Wildon
    Commented Nov 14 at 19:32
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    $\begingroup$ The Klein group in my example is a transitive subgroup of $S_4$ of order $4$ in which, as you expected, the stabiliser of every point is trivial. $\endgroup$
    – Mark Wildon
    Commented Nov 15 at 17:37
  • $\begingroup$ @MarkWildso: Thanks, I see. I will change the answer. $\endgroup$
    – Yanlong Hao
    Commented Nov 15 at 17:47
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    $\begingroup$ In other words, every power $\sigma$ of this cycle enjoys the equation $\sigma(1)+\sigma(4)=\sigma(2)+\sigma(3)$, thus the sum of the first and the fourth columns equals to the sum of the second and the third columns of the matrix $\endgroup$
    – Fedor Petrov
    Commented Nov 15 at 18:28
  • $\begingroup$ @FedorPetrov: Yes, good point. Your observation is useful for providing more counterexamples. The idea of the cyclic matrix is more useful to show in some cases, the matrix is always invertible. For example, it is easy to see for $n$ a prime number, the matrix is always invertible. In this case, the group must be generated by a $n$-cycle by considering the order of an element. And the minimal polynomials of $n$-th root of unity are $x-1$ or $\frac{x^n-1}{x-1}$. And it is not a hard exercise to very find they are not a factor for any $f(x)$ constructed as above. $\endgroup$
    – Yanlong Hao
    Commented Nov 15 at 19:17

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