Let \(G=\{g_1,\ldots,g_k\}\) be a finite group. Introduce variables \(x_{g_1},\ldots,x_{g_k}\) and consider the \(k\times k\) matrix \[ X=\bigl[x_{g_i g_j^{-1}}\bigr]. \] Let \(X=\sum_{i=1}^k A(g_i)x_{g_i}\) so that we can define a map \(g_i\mapsto A(g_i)\).
Show that \(A\) is the left regular representation of \(G\).
Show that \[ \det(X)=\prod_{\nu=1}^{h}\det\Biggl(\sum_{g\in G}A^{(\nu)}(g)x_g\Biggr)^{n_\nu} \] where \(A^{(1)},\ldots,A^{(h)}\) are a complete set of representatives of the irreducible representations of \(G\) and \(n_\nu=\dim(A^{(\nu)})\) for \(\nu=1,\ldots,h\).
Use part (b) to show that \[ \det\begin{bmatrix} x_0 & x_1 & x_2 & \cdots & x_{n-1}\\ x_{n-1} & x_0 & x_1 & \cdots & x_{n-2}\\ x_{n-2} & x_{n-1} & x_0 & \cdots & x_{n-3}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ x_1 & x_2 & x_3 & \cdots & x_0 \end{bmatrix} =\prod_{r=0}^{n-1}\bigl(x_0+\epsilon^r x_1+\epsilon^{2r}x_2+\cdots+\epsilon^{(n-1)r}x_{n-1}\bigr) \] where \(\epsilon=e^{2\pi i/n}\).
Solution
For each \(g\in G\), let \(A(g)\) denote the coefficient of \(x_g\) in \[ X=[x_{g_i g_j^{-1}}]. \] Thus \[ A(g)_{ij}= \begin{cases} 1,&g_i g_j^{-1}=g,\\ 0,&\text{otherwise}. \end{cases} \]
1. The map \(g\mapsto A(g)\) is the left regular representation of \(G\) on the vector space with basis \(e_{g_1},\ldots,e_{g_k}\).
For a fixed column \(j\), there is a unique index \(i\) such that \[ g_i=gg_j. \] By the definition above, this is exactly the unique row for which \(A(g)_{ij}=1\). Hence \[ A(g)e_{g_j}=e_{gg_j}. \] This is left multiplication by \(g\) on the basis indexed by \(G\). Therefore \[ A(g)A(h)e_{g_j}=A(g)e_{hg_j}=e_{ghg_j}=A(gh)e_{g_j} \] for every basis vector, so \(A(g)A(h)=A(gh)\) and \(A(e)=I\). Thus \(A\) is precisely the left regular representation. This proves part (a).
2. Over \(\mathbb C\), the regular representation decomposes as\[ \mathbb C[G]\cong\bigoplus_{\nu=1}^h n_\nu V_\nu, \qquad n_\nu=\dim V_\nu, \]where \(V_1,\ldots,V_h\) are representatives of the irreducible \(G\)-modules.
Maschke’s theorem makes the regular representation completely reducible. Let \(\chi_{\mathrm{reg}}\) be its character. Then \[ \chi_{\mathrm{reg}}(e)=|G|, \qquad \chi_{\mathrm{reg}}(g)=0\quad(g\ne e), \] because left multiplication by \(g\ne e\) permutes the basis \(\{e_h:h\in G\}\) without fixed points.
If \(\chi_\nu\) is the irreducible character of \(V_\nu\), its multiplicity in the regular representation is \[ \langle\chi_{\mathrm{reg}},\chi_\nu\rangle =\frac1{|G|}\sum_{g\in G}\chi_{\mathrm{reg}}(g)\overline{\chi_\nu(g)} =\overline{\chi_\nu(e)} =\dim V_\nu=n_\nu. \] This gives the asserted decomposition.
3. There is an invertible matrix \(P\) such that, for every \(g\in G\),\[ P^{-1}A(g)P = \bigoplus_{\nu=1}^h \underbrace{A^{(\nu)}(g)\oplus\cdots\oplus A^{(\nu)}(g)}_{n_\nu\text{ copies}}. \]
2. In that basis, the regular representation acts on each of the \(n_\nu\) copies of \(V_\nu\) by the irreducible representation \(A^{(\nu)}\). The change-of-basis matrix from the original regular basis to this adapted basis is the required \(P\).
4. Therefore\[ \det(X) = \prod_{\nu=1}^h \det\Bigl(\sum_{g\in G}A^{(\nu)}(g)x_g\Bigr)^{n_\nu}. \]
Since \[ X=\sum_{g\in G}A(g)x_g, \] 3 gives \[ P^{-1}XP = \bigoplus_{\nu=1}^h \underbrace{ \Bigl(\sum_{g\in G}A^{(\nu)}(g)x_g\Bigr) \oplus\cdots\oplus \Bigl(\sum_{g\in G}A^{(\nu)}(g)x_g\Bigr) }_{n_\nu\text{ copies}}. \] Determinant is invariant under similarity and multiplicative on block-diagonal matrices, so taking determinants gives exactly the displayed formula. This proves part (b).
5. Let \(G=C_n=\langle a\rangle\) and put\[ x_j=x_{a^j}\qquad(0\le j<n). \]The irreducible complex representations of \(C_n\) are the one-dimensional characters\[ \chi_r(a^j)=\epsilon^{rj}, \qquad 0\le r<n, \]where \(\epsilon=e^{2\pi i/n}\).
Since \(C_n\) is finite abelian, every irreducible complex representation is one-dimensional. A character is determined by the image of \(a\), which must be an \(n\)th root of unity. The \(n\) choices are \(\epsilon^r\), giving exactly the characters displayed above.
6. If the regular basis is ordered by\[ 1,a^{-1},a^{-2},\ldots,a^{-(n-1)}, \]then the group-determinant matrix \([x_{g_i g_j^{-1}}]\) is exactly\[ \begin{bmatrix} x_0 & x_1 & x_2 & \cdots & x_{n-1}\\ x_{n-1} & x_0 & x_1 & \cdots & x_{n-2}\\ x_{n-2} & x_{n-1} & x_0 & \cdots & x_{n-3}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ x_1 & x_2 & x_3 & \cdots & x_0 \end{bmatrix}. \]
With indices \(i,j\in\{0,\ldots,n-1\}\) and \(g_i=a^{-i}\), \[ g_i g_j^{-1}=a^{-i}a^j=a^{j-i}. \] Thus the \((i,j)\)-entry is \(x_{j-i\bmod n}\), which is exactly the displayed circulant matrix.
7. Its determinant is\[ \prod_{r=0}^{n-1} \bigl(x_0+\epsilon^r x_1+\epsilon^{2r}x_2+\cdots+\epsilon^{(n-1)r}x_{n-1}\bigr). \]
For \(C_n\), every irreducible representation has dimension \(1\), so 4 becomes \[ \det(X) =\prod_{r=0}^{n-1}\left(\sum_{j=0}^{n-1}\chi_r(a^j)x_j\right). \] Using 5, \[ \sum_{j=0}^{n-1}\chi_r(a^j)x_j =\sum_{j=0}^{n-1}\epsilon^{rj}x_j =x_0+\epsilon^r x_1+\cdots+\epsilon^{(n-1)r}x_{n-1}. \] Substitution yields the required formula, proving part (c).