Induced representations; an example

Posted: February 23, 2011 in Representations of Finite Groups
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We’re now going to use induced representations to find m degree two representations for D_{2m}, the dihedral group of order 2m. We already gave these representations in Example 1 (also see the remark after that!) but we didn’t explain how we got them!

Example. Let g_1,g_2 be the generators of D_{2m}, where o(g_1)=2, \ o(g_2)=m and g_1g_2=g_2^{-1}g_1. Let H = \langle g_2 \rangle. Show that every induced representation from H to G has degree two and find all of them.

Solution. Since H is a cyclic group, it has m representations and all of them have degree one (see Example 1). Thus, by the theorem in this post, if \rho is a representation of H, then \deg \overline{\rho}=[D_{2m}:H] \deg \rho=2 \times 1=2. Now let \rho : H \longrightarrow \text{GL}(\mathbb{C}) be a representation of H. Then \rho(g_2)=\zeta, where \zeta is an m-th root of unity. Let \overline{\rho} = \text{Ind}_H^G \rho : G \longrightarrow \text{GL}(V) be the induced representation of \rho. So V = \mathbb{C}[G] \otimes_{\mathbb{C}[H]} \mathbb{C} \cong \mathbb{C}^2, where \mathbb{C} in the tensor product is considered as a left \mathbb{C}[H]-module and the action is defined by \rho, i.e. rw=\rho(r)(w), for all r \in \mathbb{C}[H] and w \in \mathbb{C}. Now, \{H,g_1H\} is a set of all left cosets of H in G. Thus if we let v_1=1 \otimes_{\mathbb{C}[H]} 1 and v_2=g_1 \otimes_{\mathbb{C}[H]} 1, then \mathcal{B}=\{v_1,v_2\} is a \mathbb{C}-basis for V. Note that \mathcal{B} corresponds to the standard basis of \mathbb{C}^2. Now, to find \overline{\rho}, I just need to find, for all g \in G, the matrix of \overline{\rho}(g) with respect to \mathcal{B}. We know that an element g \in G is either in the form g_1g_2^j, \ 0 \leq j < m, or in the form g_2^j, \ 0 \leq j < m. So we will consider two cases:

Case 1 . g = g_1g_2^j, \ 0 \leq j < m. In order to find the matrix of \overline{\rho}(g) with respect to \mathcal{B}, we only need to find \overline{\rho}(g)(v_i), \ i=1,2. We have

\overline{\rho}(g)(v_1) = g \otimes_{\mathbb{C}[H]}1=g_1 \otimes_{\mathbb{C}[H]}(g_2^j \cdot 1) = g_1 \otimes_{\mathbb{C}[H]} \rho(g_2^j)=g_1 \otimes_{\mathbb{C}[H]} \zeta^j=\zeta^jv_2.

Similarly, since gg_1=g_2^{-j}, we have

\overline{\rho}(g)(v_2)=g_2^{-j} \otimes_{\mathbb{C}[H]} 1 = 1 \otimes_{\mathbb{C}[H]} (g_2^{-j} \cdot 1)=1 \otimes_{\mathbb{C}[H]} \rho(g_2^{-j})=\zeta^{-j}v_1.

So in this case the matrix of \overline{\rho}(g) with respect to \mathcal{B} is  \begin{pmatrix} 0 & \zeta^{-j} \\ \zeta^j & 0 \end{pmatrix}.

Case 2 . g=g_2^j, \ 0 \leq j < m. A similar argument as case 1 shows that the matrix in this case is \begin{pmatrix} \zeta^j & 0 \\ 0 & \zeta^{-j} \end{pmatrix}. \ \Box

In case 1 (and case 2) in the above, the reason that we were allowed to move powers of g_2 to the right side of tensor product is that the tensor product is over \mathbb{C}[H] and any power of g_2 is in \mathbb{C}[H] because \langle g_2 \rangle =H \subset \mathbb{C}[H].

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