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!!!!