Induced representation of $G=\operatorname{SL}_2$ by $\chi _ w$ is irreducible if $w^2\neq 1$. This is the question from Serre's book #7.4. $G=\operatorname{SL}_2(k)$, where $k$ is a finite field and $H\leq G$ such that $H$ consists of matrices $\begin{pmatrix} a & b \\ 0 & d  \end{pmatrix}$. Let $w:k^* \rightarrow \mathbb{C}^*$ be a homomorphism and $\chi_w$ be the character of degree 1 of $H$ defined by $\chi_w \begin{pmatrix} a & b \\ c & d  \end{pmatrix}=w(a)$. Show that the representation of $G$ induced by $\chi_w$ is irreducible if $w^2\neq 1$.
How can i start this problem?
 A: Compute the inner product of the induction with itself, using Frobenius reciprocity and Mackey's formula. For the latter, begin by showing that the set of double cosets $H\backslash SL_2(k)/H$ has size 2, with representatives $\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$.
A: Sorry to resurrect such an old thread, but in the interests of completeness, we provide a complete solution.
Choose $X = \begin{pmatrix} 
a & b \\
c & d 
\end{pmatrix} \in G \setminus H$. A routine calculation gets us$$A := \begin{pmatrix} 
a & b \\
c & d 
\end{pmatrix}\begin{pmatrix} 
x & y \\
0 & z 
\end{pmatrix}\begin{pmatrix} 
a & b \\
c & d 
\end{pmatrix}^{-1} = \begin{pmatrix} 
adx - acy - bcz & a(bz + ay-bx) \\
c(dx - cy - dz) & adz + acy - bcx 
\end{pmatrix}.$$This is in $H$ exactly when $y = {{d(z-x)}\over{c}}$. In this case, we can rewrite the above matrix as $$A = \begin{pmatrix} 
z & a(bz + ay - bx) \\
0 & x 
\end{pmatrix},$$ so this describes all matrices in $H_X$. Also, $$A^{-1} = \begin{pmatrix} 
x & a(bz + ay - bx) \\
0 & z
\end{pmatrix}.$$ Because $x = z^{-1}$, each matrix in $H_X$ is completely described by its top left entry. Now we can compute$$|H_X|\langle\rho^X,\text{Res}_X(\rho)\rangle_{H_X} = \sum_{Y \in H_X} \chi_w(Y^{-1})\chi_w(X^{-1}YX) = \sum_{x \in k^\times} w(x)w(x).$$Because $k$ is a finite field, $k^\times$ is a cyclic group, hence $w$ maps elements of $k^\times$ to roots of unity. Because $w^2 \neq 1$, a generator of $k^\times$ is mapped to an $n$th root of unity with $n >2$. Therefore, the last sum above cycles through all $(n/2)$th roots of unity, and each one appears the same amount of time. In particular, the sum is $0$, hence $\rho^X$ and $\text{Res}_X(\rho)$ are disjoint. Because $\chi_w$ is a degree $1$ representation, it is irreducible. Thus, by Proposition 23 of Serre's Linear Representations of Finite Groups, the induced representation of $G$ is irreducible.
