Let \(G\) and \(H\) be a finite groups, \(A:G\to GL(n,\mathbb{C})\) be a representation of \(G\) and \(B:H\to GL(m,\mathbb{C})\) be a representation of \(H\).
Define the representation \(A\times B:G\times H\to GL(n\cdot m,\mathbb{C})\).
Show that if \(A\) and \(B\) are irreducible representations, then \(A\times B\) is irreducible.
Show that every irreducible representation of \(G\times H\) is of the form \(A\times B\) where \(A\) is an irreducible representation of \(G\) and \(B\) is an irreducible representation of \(H\).
Solution
1. Let \(V=\mathbb C^n\) and \(W=\mathbb C^m\) be the representation spaces of \(A\) and \(B\).Define\[ (A\times B)(g,h)=A(g)\otimes B(h) \]on \(V\otimes_{\mathbb C}W\).
For \((g_1,h_1),(g_2,h_2)\in G\times H\), \[ \begin{aligned} (A\times B)(g_1g_2,h_1h_2) &=A(g_1g_2)\otimes B(h_1h_2)\\ &=(A(g_1)A(g_2))\otimes(B(h_1)B(h_2))\\ &=(A(g_1)\otimes B(h_1))(A(g_2)\otimes B(h_2)). \end{aligned} \] Also \((A\times B)(1,1)=I_V\otimes I_W\), so this is a representation of dimension \(nm\).
2. Its character is\[ \chi_{A\times B}(g,h)=\chi_A(g)\chi_B(h). \]
For square matrices \(X,Y\) one has \[ \operatorname{tr}(X\otimes Y)=\operatorname{tr}(X)\operatorname{tr}(Y). \] Apply this with \(X=A(g)\) and \(Y=B(h)\).
3. If \(A\) and \(B\) are irreducible, then \(A\times B\) is irreducible.
Over \(\mathbb C\), a finite-group representation is irreducible iff the inner product of its character with itself is \(1\). Using 2, \[ \begin{aligned} \langle\chi_{A\times B},\chi_{A\times B}\rangle_{G\times H} &=\frac1{|G||H|}\sum_{g\in G}\sum_{h\in H} |\chi_A(g)|^2|\chi_B(h)|^2\\ &=\langle\chi_A,\chi_A\rangle_G \langle\chi_B,\chi_B\rangle_H\\ &=1. \end{aligned} \] Thus \(A\times B\) is irreducible.
4. If \(A_1,A_2\) are irreducible representations of \(G\) and \(B_1,B_2\) are irreducible representations of \(H\), then\[ \langle\chi_{A_1\times B_1},\chi_{A_2\times B_2}\rangle_{G\times H} = \langle\chi_{A_1},\chi_{A_2}\rangle_G \langle\chi_{B_1},\chi_{B_2}\rangle_H. \]
The same separation of the double sum used in 3 gives the displayed equality.
5. Therefore the representations \(A\times B\), with \(A\in\operatorname{Irr}(G)\) and \(B\in\operatorname{Irr}(H)\), are pairwise nonisomorphic irreducible representations of \(G\times H\).
By 3 each is irreducible. By 4 and orthogonality of irreducible characters, the inner product is \(1\) exactly when \(A_1\cong A_2\) and \(B_1\cong B_2\), and is \(0\) otherwise.
6. These external products exhaust all irreducible representations of \(G\times H\).
Let \(k(K)\) denote the number of conjugacy classes of a finite group \(K\). Two pairs \((g,h)\) and \((g',h')\) are conjugate in \(G\times H\) iff \(g\) is conjugate to \(g'\) in \(G\) and \(h\) is conjugate to \(h'\) in \(H\). Hence \[ k(G\times H)=k(G)k(H). \] For every finite group over \(\mathbb C\), the number of irreducible representations equals the number of conjugacy classes. Therefore \[ |\operatorname{Irr}(G\times H)| =k(G\times H) =k(G)k(H) =|\operatorname{Irr}(G)|\,|\operatorname{Irr}(H)|. \] By 5 we already have exactly this many pairwise nonisomorphic irreducible external products. Thus every irreducible representation of \(G\times H\) is of the form \(A\times B\).