## Complete set of irreducible representations of small groups (2)

Posted: February 10, 2011 in Representations of Finite Groups
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Example 3. Irreducible Representations of $D_8$ : In part 5) in here we showed that $D_8,$ the dihedral group of order $8,$ has five non-equivalent irreducible representations. Four of them have degree one and one of them has degree two. So let $D_8 = \langle g_1, g_2 : \ g_1^2=g_2^4=(g_1g_2)^2=1 \rangle.$ Every element of $D_8$ can be written uniquely as $g_1^j g_2^k,$ where $0 \leq j \leq 1$ and $0 \leq k \leq 3.$

1) Representations of degree one. These have been completely described in Example 2 in this post. In our case $m=4,$ which is an even number. So there are four representations of degree one, say $\rho_1, \rho_2, \rho_3, \rho_4.$ They are defined on $D_8$ by

$\rho_1(g)=1, \ \rho_2(g)=(-1)^k, \ \rho_3(g)= (-1)^j, \ \rho_4(g)= (-1)^{j+k},$

where $g = g_1^j g_2^k, \ 0 \leq j \leq 1, \ 0 \leq k \leq 3.$

2) Representation of degree two. Here we described $m$ representations of degree two for the dihedral group $D_{2m}.$ We proved there that all except those corresponding to $\zeta = \pm 1, \ m \geq 3,$ $\rho_0$ are irreducible. In our case $m=4.$ So there are two irreducible representations and we’ll pick the one corresponding to $\zeta = i.$ Let’s call it $\rho.$ So if  $v \in \mathbb{C}^2,$ then $\rho$ is defined on $D_8$ by

$\rho(g_1^jg_2^k)(v)= \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}^j \begin{pmatrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{pmatrix}^k v,$

for all $0 \leq j \leq 1$ and $0 \leq k \leq 3.$ I’ll leave it to the reader to find an explicit formula for $\rho(g_1^j g_2^k)(v).$

Example 4. Irreducible Representations of $Q_8$ : In part 4) in here we showed that $Q_8,$ the quaternion group of order $8,$ has five non-equivalent irreducible representations. Four of them have degree one and one of them has degree two.

1) Representations of degree one. These were fully described in Example 1 in this post. We showed that if $\rho$ is a degree  one representation of $Q_8,$ then $\rho(1)=\rho(-1)=1, \ \rho(i)=\pm 1, \ \rho(j)=\pm 1.$
Therefore $\rho(-i)=\rho(-1)\rho(i)=\rho(i)$ and similarly $\rho(-j)=\rho(j).$ Finally, since $k=ij,$ we  have $\rho(k)=\rho(i)\rho(j)$ and $\rho(-k)=\rho(k).$

2) The representation of degree two. So we need $2 \times 2$ matrices with entries in $\mathbb{C}$ which satisfy the same relations as $i,j, k$ do. Let $A = \begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$ Let $C=AB.$ Now if $v \in \mathbb{C}^2,$ we define $\rho(i)(v)=Av$ and $\rho(j)(v)=Bv.$ Since $A,B$ satisfy the same relations as $i,j$ do, $\rho$ will give us a group homomorphism from $Q_8$ to $\text{GL}(\mathbb{C}^2)$ and hence it is a representation of degree two. So, for example $\rho(-1)(v)=\rho(i^2)(v)=\rho(i) \rho(i)(v)=A^2v,$ etc. It is easy to see that $A,B$ have no common eigenvector and thus $\rho$ is irreducible by the theorem we proved in here.