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Let $G$ be a finite group with elements $g_1, g_2, \ldots, g_n$. We define the group matrix by $$X_G = [x_{g_ig_j^{-1}}].$$

We then can define the group determinant as $$\det X_G = \Theta_G.$$

$\Theta_G$ will be a polynomial over the $x_{g_k}$. It seems like the group determinant would be independent (perhaps just up to sign $(\pm)$) from the indexing of the elements of $G$ since any renumbering of the elements would just cause row and column swaps.

Is this a proper way to think of this?

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  • $\begingroup$ Maybe this way $(\det(A) = \sum\limits_{\sigma \in S_n} \hbox{sgn}(\sigma) \prod\limits_{i=1}^n a_{i,\sigma_i})$ is more useful here, imho. $\endgroup$ May 11 '15 at 23:52
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The determinant is indeed independent from the indexing of the elements.

One way to see this is the following. Suppose you index your elements in a different way; this means you take an element $\sigma$ of the symmetric group $S_n$, and you order your element as $g_{\sigma(1)}, g_{\sigma(2)}, \ldots, g_{\sigma(n)}$. With this new ordering, the matrix becomes $$X^\sigma_G=[x_{g_{\sigma(i)}g_{\sigma(j)}^{-1}}]. $$

Now, let $P$ be the permutation matrix associated to $\sigma$, that is, $P=[\delta_{\sigma(i), j}]$, where $\delta_{k, \ell}$ is $1$ if $k=\ell$ and $0$ else. If $M$ is any $n\times n$ matrix, then the matrix $PM$ is obtained by reordering the rows of $M$ by applying $\sigma$; similarly, $MP^{-1}$ is obtained by reordering the columns of $M$ by applying $\sigma$. Combining these two observations, you get that $X^\sigma_G = PX_G P^{-1}$. Taking the determinant, you have $det(X^\sigma_G) = det(X_G)$.

Note that the determinants are exaclty equal, not up to a sign.

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