What is known about the algebraic variety $V\_G$ defined by $det(X\_G) = 1$ where $X\_G$ is the group matrix $(x\_{g\_ig\_j^-1})$ of a finite group $G$? It is known that two finite groups having the same determinant are isomorphic.


**Edit:** Are there any results concernig the group structure on $V\_G$?



What is known about the algebraic variety defined by the group determinant?
Per the suggestion of Neil Strickland, here is an answer. There are some excellent [notes](http://www.math.uconn.edu/~kconrad/articles/groupdet.pdf) of Keith Conrad about all of this.


If $k$ is an algebraically closed field of characteristic prime to $n$, the order of $G$, then classical results about representations of finite groups imply an isomorphism of associative, unital $k$-algebras, $$\phi:k[G]\to A\_1\times \dots \times A\_r,$$ between the group ring $k[G]$ is isomorphic and a product $A\_1\times \dots \times A\_r$ of matrix algebras $A\_i \cong \text{Mat}\_{n\_i\times n\_i}(k)$ (if $k$ is not algebraically closed, each $A\_i$ may be a matrix algebra over a division ring). Here the integers $n\_1\leq \dots \leq n\_r$ are the ranks of the finitely many (non-isomorphic) irreducible representations of $G$ over $k$. In particular, because there is a trivial one-dimensional represenation, $n\_1$ equals $1$.


For any associative, unital $k$-algebra $A$,
multiplication on the right in $k[G]$ defines a "regular representation", i.e., a homomorphism of associative, unital $k$-algebras, $$ \rho\_A: A \to \text{Hom}\_k(A,A),\ a \mapsto (b\mapsto b\cdot a).$$ If $A$ is a finite dimensional $k$-vector space, then there is the standard determinant polynomial on $\text{Hom}\_k(A,A)$. The composition with $\rho\_A$ is the algebra norm, $\Theta\_A$, which is some power of the reduced norm. 


For $A=k[G]$, for the standard $k$-basis of $k[G]$, $(\mathbf{b}\_\gamma)\_{\gamma \in G}$, and for a general element $a = \sum\_{\gamma} x\_\gamma \mathbf{b}\_\gamma$ of $k[G]$, the matrix entry of $\rho(a)$ with respect to $\mathbf{b}\_\gamma$ (in the domain of $\rho(a)$) and $\mathbf{b}\_\delta$ (in the target of $\rho(a)$) is $$ \rho(a)\_{\delta,\gamma} = x\_{\delta\cdot \gamma^{-1}}.$$ Thus the determinant $\text{det}(x\_{\delta\cdot \gamma^{-1}})\_{\gamma,\delta}$ equals $\Theta\_G(a)$, i.e., composition of $\rho$ with the usual determinant on $\text{Hom}\_k(A,A) \cong \text{Mat}\_{n\times n}(k)$. 


The algebra norm is compatible with isomorphism of associative, unital $k$-algebras. Thus, via $\phi$, $\Theta\_G$ is the same as $\Theta\_{A\_1\times \dots \times A\_r}$. Also the algebra norm diagonalizes, i.e., for $(a\_1,\dots,a\_r)\in A\_1\times\dots \times A\_r$, $$\Theta\_{A\_1\times \dots \times A\_r}(a\_1,\dots,a\_r) = \Theta\_{A\_1}(a\_1) \cdots \Theta\_{A\_r}(a\_r).$$ Finally, $\Theta\_{\text{Mat}\_{n\times n}(k)}$ equals $\text{det}^n$, where $\text{det}$ is the usual determinant polynomial on $\text{Mat}\_{n\times n}(k)$. Now consider the multiplicative morphism,
$$ \text{det}:A\_1\times \dots \times A\_r \to k^r, \ (a\_1,\dots,a\_r) \mapsto (\text{det}(a\_1),\dots,\text{det}(a\_r)).$$ The algebra norm is the composite of this morphism with the morphism, $$k^r \mapsto k, \ (t\_1,\dots,t\_r) \mapsto t\_1^{n\_1}\cdots t\_r^{n\_r}.$$ Since $n\_1$ equals $1$, for every $t\in \mathbb{G}\_m$, the fiber of $\Theta\_{k[G]}$ over $t$ is isomorphic to $\textbf{GL}(A\_2)\times \dots \times \textbf{GL}(A\_r)$ via the isomorphism, $$\textbf{GL}(A\_2)\times \dots \times \textbf{GL}(A\_r) \to \Theta\_{k[G]}^{-1}(t), \ (a\_2,\dots,a\_r) \mapsto (t\cdot\text{det}(a\_2)^{-n\_2}\cdots \text{det}(a\_r)^{-n\_r}, a\_2,\dots,a\_r).$$ Thus, the fiber of $\Theta\_{k[G]}$ over $1$ is isomorphic, as a $k$-group scheme, to the product $\textbf{GL}(A\_2)\times \dots \times \text{GL}(A\_r)$. 


In particular, the structure of $\text{Ker}(\Theta\_{k[G]})$ as a $k$-group scheme is equivalent to the information of the sequence of integers $(n\_1=1,n\_2,\dots,n\_r)$ of ranks of the irreducible $k$-representations of $G$. 


