Clarifications on the faithful irreducible representations of the dihedral groups over finite fields. I would like to clarify a few things in the answer to this question:
Faithful irreducible representations of cyclic and dihedral groups over finite fields
1) When a representation extends, what does the remaining generator map to under the representation?
2) Why does a representation extend if and only if $z^{-1} = z^{p^{d}}$ for some $d$ with $1 \leq d \leq e$? (It appears this is to do with $z$ being $\mathbb{F}_p$-conjugate to $z^{-1}$?)
3) How do we know these are faithful, irreducible representations?
4) How do we know these are the only faithful, irreducible representations?
Many thanks.
 A: I think you need to make this question more stand alone. Foe example, 1 doesn't make much sense on its own. For 2, you need to understand some Clifford theory. An irreducible representation over $F_{p}$ of a cyclic group $\langle z \rangle$ of order $n$ becomes over the algebraic closure, a direct sum of distinct absolutely irreducible representations which are conjugate via the Frobenius automorphism $x \to x^{p}.$ So you need to look at the action of this automorphism on the $1$-dimensional representations of $\langle z \rangle.$ Since $z$ is conjugate to $z^{-1}$ within the dihedral group, any $F_{p}$ representation of the cyclic group which is to have any chance of extending to the dihedral group has to have  $1$-dimensional representation and its dual in the same orbit under that Frobenius automorphism.
When you think about what this means, you get condition 2 ( strictly, we have only demonstrated necessity here, but sufficiency is similar).
For 3,4  Any faithful representation of the dihedral group has to "lie over" a faithful representation 
of the cyclic subgroup of index $2.$ On the other hand, if it does lie over such a faithful representation, then since every element outside the cyclic subgroup has order $2$ and inverts every element of the cyclic subgroup, no element outside the cyclic subgroup can lie in the kernel of the representation, while no non-identity element of the cyclic subgroup is in the kernel, so only the identity is in the kernel.
