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I have some questions about this post:

Faithful irreducible representations of cyclic and dihedral groups over finite fields

I would appreciate it really if someone could help me.

1) Do I get with this method really just the $\underline{faithful}$ irreducible representations? So there can be still other irreducible representation which are not faithful (and have dimension $>1$). Can one count them too?

2) If an irreducible representation of degree e of the normal subgroup $⟨z⟩$ extends to the whole group, this does mean, that it still has the same degree $e$, not?

3) Where do the condition

" it extends, if $z^{-1}=z^{p^d}$ for some $d$ with $1≤d≤e$ "

come from. Is it a theorem in Clifford's theory or just a conclusion, which you can see, if you work long enough with it? My book does not mention this.

[edit] Could a member please accept the answer. I cannot comment on my own post or accept the answer after 60min.

tinlyx
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Haruki S.
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1 Answers1

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  1. If you consider irreducible representations of a dihedral group, and you restrict to the cyclic subgroup of index $2,$ the restriction need not be faithful. But the point is that if it isn't, you are really dealing with a representation of a smaller dihedral group. So if you work inductively, you only really need to consider the faithful case.
  2. If we speak of a representation of a subgroup $H$ extending to the whole group, this means by definition that the degree remains unchanged.
  3. This is a consequence of Clifford's theorem and some understanding about Galois conjugates. It is not an immediate consequence of Clifford's theorem. The point is that if we have a representation $\sigma$ of the group $\langle z \rangle$ over the field of $p$ elements, then it is a sum of Galois conjugate absolutely irreducible representations. If $\tau$ is one of these, then $z \to \tau(z^{p^k})$ is another representation. The representation $\sigma$ will be equivalent to its dual representation if and only if the dual representation of $\tau$ is one of these Galois conjugates using powers of the Frobenius automorphism. That is, if and only if $\tau(z^{-1}) = \tau(z^{p^k})$ for some $k.$
Geoff Robinson
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