Let $k$ be a finite field of characteristic $p\neq 2$ (in fact, one only needs to consider the case $p\in\{3,5\}$), let $\Sigma$ be a finite set of primes containing $\infty$ and $p$, and
$$\rho_{0}:{\rm Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})\rightarrow {\rm GL}_{2}(k)$$
an absolutely irreducible representation, meaning $\rho_{0}\otimes\overline{k}$ cannot be written as the direct sum of two one-dimensional subrepresentations, where $\mathbb{Q}_{\Sigma}$ is the largest Galois extension of $\mathbb{Q}$ unramified outside the primes in $\Sigma$. Assume that if $\tau\in\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})$ is complex conjugation, then $\det(\rho_{0}(\tau))=-1$.
Then $\rho$ induces a projective representation
$$\tilde{\rho}_{0}:\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})\longrightarrow\operatorname{PGL}_{2}(k).$$
Assume this projective representation has dihedral image, meaning
$$\operatorname{image}(\tilde{\rho}_{0})\cong\left<s,r\mid s^{2}=r^{m}=(sr)^{2}=1\right>$$
for some $m\in\mathbb{N}$, and assume further that $\rho_{0}|_{\mathbb{Q}(\sqrt{(-1)^{\frac{p-1}{2}}p})}$ is absolutely irreducible.
The action of $\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})$ on $k^{2}$ induces an action on $V_{\lambda}=\operatorname{Hom}(k^{2},k^{2})$, namely by conjugation. (What that $\lambda$ stands for is of no significance here.) Since $p\neq 2$, one has a direct sum of $k[\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})]$-modules
$$V_{\lambda}=W_{\lambda}\oplus k\text{,}$$
where $W_{\lambda}$ denotes the space of $\operatorname{trace}$–$0$ matrices and $k$ is the space of scalar multiplications.
Let $K_{1}$ be the splitting field of $\rho_{0}$ (i.e. $\operatorname{Gal}(\mathbb{Q}_{\Sigma}/K_{1})=\operatorname{ker}(\rho_{0})$), and
$$G:=\operatorname{Gal}(K_{1}/\mathbb{Q})=\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})/\operatorname{ker}(\rho_{0})=\operatorname{image}(\rho_{0}).$$
Since $\overline{\rho}_{0}$ has dihedral image, $\rho_{0}\otimes\overline{k}=\operatorname{Ind}_{H}^{G}(\chi)$ for some character $\chi$.
Question: Wiles now makes the following claims.
(1.) Under the above conditions, $W_{\lambda}\otimes\overline{k}=\delta\otimes\operatorname{Ind}_{H}^{G}(\chi/\chi')$ where $\chi'$ is the quadratic twist of $\chi$ by any element of $G\setminus H$ and $\delta$ is the quadratic character $G\longrightarrow G/H$ (what does that even mean – $W_{\lambda}\otimes\overline{k}$ decomposes as a quadratic character + something else?)
(2.) since $M(\zeta_{p^{n}})$ is Abelian over $\mathbb{Q}$, where $\mathbb{Q}\subseteq M\subseteq K_{1}$ such that $G/H=\operatorname{Gal}(M/\mathbb{Q})$, one always finds for any $n\in\mathbb{N}$ an $x\in\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ which fixes $\mathbb{Q}(\zeta_{p^{n}})$ and $\tilde{\rho}_{0}(x)\neq 1$ as long as $m\neq 2$ (i.e. $\operatorname{image}(\rho_{0})\neq\mathbb{Z}/2\times\mathbb{Z}/2$).
Why is that the case? Maybe I'm not seeing the wood for all the trees here, but who knows. EDIT: Also, should it not be $\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})$ instead of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$?
Best Answer
(1): First, let's define some notations. Let's call $G$ the Galois group, $D$ its dihedral image, $C$ its distinguished cyclic subgroup, $H$ its inverse image in $G$. We fix some $s \in G$ whose image in $D$ is the $s$ of the dihedral presentation.
The quotient of $H$ by its scalar matrices (a central subgroup) is $C$ cyclic, so that $H$ is abelian. Then, we can note that $\rho_0$ is given on $H$ by direct sum of two characters $\chi$ and $\chi'$.
We change the basis for $\rho_0$ so that $k(1,0)$, $k(0,1)$ are the lines stable under $H$ (with characters $\chi,\chi'$) and $\rho_0(s)=\begin{bmatrix}0 & 1\\1&0\end{bmatrix}$.
Then we can easily see that $V = kI_2 \oplus k\cdot \mathrm{diag}(1,-1) \oplus A$, where $A$ is the space of antidiagonal matrices. Clearly, $G$ acts trivially on $kI_2$, $H$ acts trivially on the second summand, but $s$ acts by $-1$, so it is the quadratic character $G \rightarrow G/H$.
As for the last summand, it has a canonical basis ($\begin{bmatrix}0 & 1\\0&0\end{bmatrix}$ and its transpose) that $s$ permutes, and $\mathrm{diag}(a,b)$ acts by $\mathrm{diag}(a/b,b/a)$ on this basis, and we recognize the induced representation from $H$ to $G$ by $\chi/\chi'$ (provided that the twist of $\chi/\chi'$ is $\chi'/\chi$, which mostly follows from the dihedral presentation).
(2): I'm interpreting the question as follows: assume $m \neq 2$ and let $n\geq 1$. Why is there an $x \in \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that $x$ fixes $\mathbb{Q}(\zeta_{p^n})$ and $\overline{\rho_0(x)} \neq 1$?
Answer: assume that there is no such $x$: then $G_{\mathbb{Q}(\zeta_{p^n})} \subset G_{K_2}$, where $K_2 \subset K_1$ is the subextension such that $Gal(K_1/K_2)$ is mapped by $\rho_0$ to the scalar matrices. In other words, $K_2 \subset \mathbb{Q}(\zeta_{p^n})$. In particular, $K_2/\mathbb{Q}$ is abelian. But the Galois group of $K_2/\mathbb{Q}$ is isomorphic to $D$ (ie is dihedral), so isn't abelian except if $m \leq 2$. We get a contradiction and we're done.
For the edit question, I don't know Wiles' paper very well, but as far as I understand, the goal is to prove that some Galois representations are modular using deformation theory. But usually, one needs some finiteness properties to use deformation theory (see Mazur's 1989 paper in "Galois group over Q" iirc), which are not satisfied by $G_Q=Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, but are satisfied by Galois groups with "restricted ramification" ie $Gal(\mathbb{Q}_{\Sigma}/\mathbb{Q})$.
(exercise/example: use cyclotomic extensions to show that $H^1(G_Q,\mathbb{F}_p):= \mathrm{Hom}(G_Q,\mathbb{F}_p)$ is infinite).