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Let $G$ be a finite group, $H \unlhd G$, $[G:H] = 2$, $\pi$ - irreducible complex representation of $G$, $\epsilon$ is also a representation of $G$ such that $\epsilon(g) = id_{\mathbb{C}}$ if $g \in H$ and $\epsilon(g) = -id_{\mathbb{C}}$ if $g \not \in H$.

Suppose $\pi \cong \pi \otimes \epsilon$. I need to prove the following:

  1. $Res_H^G\pi = \phi_1 \oplus \phi_2$, where $\phi_1$ and $\phi_2$ are both irreducible representations of $H$
  2. $\phi^g_1 \cong \phi_2$, where $\phi^g_1(h) = \phi_1(ghg^{-1})$ for some $g \in G\backslash H$

I managed to prove the first part of the question using some character theory (proving that the restriction can either be irreducible or a direct sum of two irreducible representations), but I can't do the second one.

I tried proving that $\phi'_1 \not \cong \phi_1$ to start with, but couldn't even do that. Any help will be appreciated.

Thanks!

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  • $\begingroup$ I think you mean you start with proving $\phi_1\not\cong\phi_2$. If $\phi_1\cong\phi_2$ then $\langle\operatorname{Res}^G_H\pi,\operatorname{Res}^G_H\pi\rangle_H=4$ not $2$. $\endgroup$ Commented Oct 9, 2023 at 20:00

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Both induction and restriction between $H$ and $G$ can be very precisely described because the index is two. This is almost certainly overkill for this particular question, but I find it clarifies things.

In what follows, let $V$ be a representation of $G$, and $W$ a representation of $H$. Let $\omega \in G \setminus H$. We write $W^\omega$ to denote the representation of $H$ obtained from $W$ by first conjugating $H$ by $\omega$.

From the formula of the character of an induced representation, one has the following two identities: $$ \text{Res}_H^G \text{Ind}_H^G W = W \oplus W^\omega, \\ \text{Ind}_H^G \text{Res}_H^G V = V \oplus (V \otimes \epsilon). $$ Coupling this together with Frobenius Reciprocity $$ (\text{Res}_H^G V, W)_H = (V, \text{Ind}_H^G W)_G $$ we have that $$ (\text{Res}_H^G V, \text{Res}_H^G V)_H = (V,V)_G + (V, V \otimes \epsilon)_G,\\ (\text{Ind}_H^G W, \text{Ind}_H^G W)_G = (W,W)_H + (W, W^\omega)_H $$

Therefore, when $V$ and $W$ are irreducible, the induction / restriction is either irreducible, or the sum of two non-isomorphic irreducible representations.

Using these identities, we can show the following claim.

Claim: Suppose that $V$ is an irreducible representation of $G$ and $W$ is an irreducible subrepresentation of $\text{Res}_H^G V$. Then there are exactly two cases that can occur:

Case 1: $V \otimes \epsilon \cong V$, $W \not\cong W^\omega$, $\text{Ind}_H^G W = V$, $\text{Res}_H^G V = W \oplus W^\omega$.

Case 2: $V \otimes \epsilon \not\cong V$, $W \cong W^\omega$, $\text{Ind}_H^G W = V \oplus (V \otimes \epsilon)$, $\text{Res}_H^G V = W$.

Proof: There are at most 4 cases, depending on whether $W \cong W^\omega$ and whether $V \cong V \otimes \epsilon$. We first show that the two pairs above are the only two options of the four.

Suppose first for a contradiction that $V \otimes \epsilon \cong V$ and $W \cong W^\omega$. Then $\text{Res}_H^G V = W \oplus U$ is the direct sum of two irreducible non-isomorphic representations. Then $$ \text{Ind}_H^G \text{Res}_H^G V = \text{Ind}_H^G W \oplus \text{Ind}_H^G U $$ and $\text{Ind}_H^G W$ is the direct sum of two non-isomorphic sub-representations. In particular, $\text{Ind}_H^G \text{Res}_H^G V$ has length $\geq 3$. However $\text{Ind}_H^G \text{Res}_H^G V = V \oplus (V \otimes \epsilon)$ has length 2, this is a contradiction.

Suppose next for a contradiction that $V \otimes \epsilon \not\cong V$ and $W \not\cong W^\omega$. Then both $\text{Ind}_H^G W$ and $\text{Res}_H^G V$ are irreducible. However this means that $\text{Res}^G_H V = W$, and so $$ \text{Ind}_H^G \text{Res}_H^G V = \text{Ind}_H^G W $$ has length $1$, which again is a contradiction.

Therefore, we are left with analysing the two cases above.

In Case 1, $\text{Ind}^G_H W$ is irreducible, and by the F.R. formula above, $V = \text{Ind}^G_H W$. Furthermore, again by this formula, $(W, \text{Res}^G_H V) = 1$, and similarly $(W^\omega, \text{Res}^G_H V) = 1$ because $\text{Ind}^G_H W = \text{Ind}^G_H W^\omega$.

In Case 2, $\text{Res}_H^G V$ is irreducible and thus $\text{Res}_H^G V = W$. By F.R. again, this means that $(V, \text{Ind}^G_H W) = 1$, and similarly again by F.R. $(V \otimes \epsilon, \text{Ind}^G_H W) = 1$ because $\text{Res}_H^G V = \text{Res}_H^G (V \otimes \epsilon)$.

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  • $\begingroup$ Thank you so much for such a detailed answer! $\endgroup$
    – Jayden
    Commented Oct 11, 2023 at 13:26

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