Is the image of $\overline{\rho_{E,p}}$ in $PGL_2$ always isomorphic to $A_5$ if $p$ does not divide the order of the image of $\rho_{E,p}$?

Question

I have the following setting: Given an elliptic curve $E$ over $\mathbf{Q}$ and $p>5$ a prime of good ordinary reduction. Let ${G}_{k}=\text{Gal}(K(E_{p^k})/K)$ with representation $\rho_{E,p^k}:G\rightarrow\text{GL}_{2}(\mathbf{Z}/{p^k}\mathbb{Z})$,where $K$ is imaginary quadratic and $p$ inert in$K/\mathbf{Q}$.
Suppose the image $\rho_{E,p}(G_{1})$ contains no element of order $p$. Is it true that
(1.) The image of the induced representation $$\overline{\rho_{E,p}}:G_1\rightarrow\text{PGL}_{2}(\mathbb{F}_{p})$$ is isomorphic to $A_5$?
(2.) $H^{1}({G}_{1},{E}_{p})=0$.
[Ciperiani & Wiles] make that claim in https://www.ma.utexas.edu/users/mirela/solvable.pdf (Prop. 1.3.1)
It escapes me why that is the case. A reference might also be good.

Answer 1

You have slightly misunderstood the situation. There is an additional assumption, namely that the image is non-solvable.

Now there is a classification of the subgroups of $\mathrm{PGL}_2(\mathbb{F}_p)$. (Probably the paper of Serre that they cite has it --- did you look at that? Otherwise, one place it is discussed is in Swinnerton-Dyer's article in LNM 350.)

My memory (but you should check) is that a subgroup either contains the image of $\mathrm{SL}_2(\mathbb F_p)$ (in which case its order is divisible by $p$), is contained in the image of a Borel (in which case it is solvable), is contained in the normalizer of a Cartan (in which case it is solvable), or is contained in $A_4,$ $S_4$, or $A_5$, the first two of which are again solvable.

Since any proper subgroup of $A_5$ is also solvable, we are reduced to either being contained in $A_5$, or containing the image of $\mathrm{SL}_2(\mathbb F_p)$, which are the two cases considered in the argument.

As for the vanishing of cohomology, isn't just a consequence of the fact that the module has $p$-power order, while the group has order prime-to-$p$ by assumption?