Computing character of a representation and irreduciblity for a finite field $k$ I have $G = SL_2(k)$ a group. $H \leq G $ and $H = \lbrace $ $\begin{bmatrix}
    a & b \\
    0 & d\\
\end{bmatrix} \vert a,b,d \in k \rbrace $
Now $\omega : k^{*} \rightarrow \mathbb{C}^{*} $ a homomorphism 
also $\chi_{\omega}$ is a 1 D character of $H$ representation given by $\chi_{\omega} (h) = \omega(a)$ where $h \in H$ 
I need to compute $\chi_{\omega}$ and also 
show that representation of $G$ induced by $\chi_{\omega}$  is irreducible if ${\omega}^2 \neq 1$ 
I don't know the modules approach. I have used only groups so far. 
 A: For Frobenius reciprocity see this link.  
In this case $H$ is the stabilizer of $(1:0)$ in the action of $G=\mathrm{SL}_2(k)$ on $\mathbb{P}^1(k)$. So if $\mathbb{P}^1(k) =\{x_0=(1:0),x_1,\ldots,x_q\}, (q = |k|)$ and for every $x_i \in \mathbb{P}^1(k)$, $g_i\in G$ be an element with $g_i(1:0) = x_i$, then $\{g_0,g_1,\dots,g_q\}$ constitutes a set of representatives for $G/H$. Now for an arbitrary $g\in G$, $g_i^{-1}gg_i\in H$ iff $x_i$ is an eigenvector of $g$ and in this case $\chi_\omega(g_i^{-1}gg_i)$ is equal to the eigenvalue of $g$ in the direction of $x_i$.
So if $\rho= \mathrm{Ind}_H^G(\chi_\omega)$and $\chi = \chi_\rho$ be the corresponding character, for an arbitrary element $g\in G$ we have this cases for $\chi(g)$:


*

*$g=\pm 1$ then $\chi(g) = \pm (q+1)$, (2 elements)

*$g\neq \pm1$ has only one eigenvalue $\pm 1$, then $g$ has only one eigenvector and $\chi(g)=\pm 1$, ($2(q+1)q$ elements)

*$g$ has two different eigenvalues $a,a^{-1}\in k$, then $\chi= \omega(a)+{\omega(a^{-1})}$, ($(q+1)(q-1)$ elements for each $a$),

*$g$ has no eigenvalues in $k$ and $\chi(g)=0$.


So we have:
$$\sum_{g\in G} |\chi(g)|^2= 2(q+1)^2 + 2(q+1)(q-1) + \frac12 q(q+1)\sum_{a\in k^*\backslash\{\pm1\}}(\omega(a)+\omega(a)^{-1})^2\\= (q+1)(q^2-q) + q(q+1)\sum_{a\in k}\omega(a)^2\\
=  ( 1 + \{1 \;\text{if } \omega^2 =1 \})\, |\mathrm{SL}_2(k)|.$$
Therefore $\rho$ is irreducible iff $\frac1G \sum|\chi(g)|^2=1$ iff $\omega^2\neq 1$.  
