## 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 $\langle g_2 \rangle =H \subset \mathbb{C}[H].$