Let H be a subgroup of index 2 in G. Let $\epsilon$ be the nontrivial character that factors through G/H:$\\$

$$\epsilon:G\rightarrow G/H \cong \{-1,1 \}\ \subset \mathbb{C}^* $$ Let $\pi$ be an irreducible representation of G. Then show that :$\\$

(1) If $\pi$ $\ncong$ $\pi \otimes \epsilon$ then $\pi|_{H}$=$\pi_1 \oplus \pi_2$, where each $\pi_{i}$ is an irreducible representation of H and $\pi_{2} \cong \pi_{1}^{g}$ $\\$

$$\pi_{1}^{g}:H \rightarrow GL(V) $$ given by $\pi_{1}^{g}(h)=\pi_{1}(g^{-1}hg)$ $\forall h \in H$. $\\$

(2) If $\pi$ $\cong$ $\pi \otimes \epsilon$, then $\pi$ restricted to H is irreducible. $\\$

My attempt: First of all let $\chi$ be the character of the representation $\pi$ of G. We know $\\$ $$\langle \chi_{H}, \chi_{H} \rangle \leq [G:H]\langle \chi,\chi \rangle$$

and equality holds iff $\chi(g)$ vanishes for all g $\in$ G-H. $\\$

In (2) by the condition we have $\chi(g) \neq \chi(g)\epsilon(g)$ implies $\chi(g) \neq 0$ for some g $\in$ G-H. Hence by above we have a strict inequality. Since $\langle \chi,\chi \rangle$=1 and [G:H]=2 we have $\langle \chi|_{H},\chi|_{H} \rangle$=1 , thereby proving (2). $\\$

Now for (1) we have $\chi(g)=\chi(g)\epsilon(g)$ and hence $\chi(g)=0$ for all g $\in$ G-H. Clearly we have equality in the above result and hence $\langle \chi|_{H},\chi|_{H} \rangle$=2. This proves that there are only two irreducible constituents of H in $\chi|_{H}$ of multiplicity 1 each. So, $\pi|_{H}$=$\pi_1 \oplus \pi_2$, where each $\pi_{i}$ is an irreducible representation of H. But then I can't prove the next assertion of (1). I hope my attempt of (2) and a partial part of (1) is correct. Thanks in advance for helping!!!!

  • $\begingroup$ In your argument for (2) you need to show that it holds for some $g\in G\setminus H$, not just some $g\in G$. For the last part of (1) you just need to show that $\pi_1^g\not\cong\pi_1$ and that further $\pi_1^g$ is a constituent of the restriction. $\endgroup$ – Tobias Kildetoft Mar 31 '17 at 8:41
  • $\begingroup$ Yeah for (2) that's fine, I actually missed that.. I have edited it now.. for the last part 1 you have a good idea. Let me try it out... Thanks!! $\endgroup$ – Riju Mar 31 '17 at 10:28
  • $\begingroup$ @TobiasKildetoft I tried your way. I have shown that $\pi_1^{g}$ is an irreducible constituent of $\pi|_{H}$ . But I can't show that $\pi_1$ and $\pi_1^{g}$ are inequivalent. I tried to show that their characters are different but couldn't do that. I know I am missing out on a small observation. Can you please help me with that. Thanks!! $\endgroup$ – Riju Mar 31 '17 at 11:47
  • $\begingroup$ I may need to think a bit more, but what happens if those are equal for all $g$? What do you get if you compute the induced character and its inner product with itself? $\endgroup$ – Tobias Kildetoft Mar 31 '17 at 12:08
  • 1
    $\begingroup$ Well, you have concluded that the induced representation, which has twice the degree of $\pi_1$, is a sum $\pi + \chi$ for some $\chi$. Since each of these has $\pi_1$ as a constituent when restricted to $H$, they must both have the same degree as $\pi_1$ since this is the only way to get that the sum of their degrees is twice the degree of $\pi_1$. But then restricting either must give you precisely $\pi_1$. $\endgroup$ – Tobias Kildetoft Mar 31 '17 at 14:17

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