Restriction of irreducible representation; index 2 subgroup

The problem is as follows:

Let $$G$$ be a finite group and $$H$$ a subgroup of index 2. Assume that whenever two elements of $$H$$ are conjugate in $$G$$, they are conjugate in $$H$$. Prove that the restriction of every irreducible complex representation of $$G$$ to $$H$$ remains irreducible.

The first part of the problem involved proving that if $$V$$ is an irreplaceable of $$G$$, then $$Res_G^H(V)$$ is either irreducible or is the direct sum of two non-isomorphic irreducible representations. For this, I used the identity $$Ind (Res (V)) = V \otimes P$$, where $$P$$ is the permutation representation of $$G$$ on $$G/H$$. Simply by dimensional arguments and Frobenius reciprocity, we have $$(\chi_{Res (U)}, \chi_{Res (U)}) = (\chi_{Ind(Res(U))}, \chi_U) = 1 \text{ or } 2$$, which gives the first part.

For the second, I was thinking of investigating $$V \otimes P$$ more. Specifically, $$P$$ breaks up into a trivial representation and a representation corresponding to the homomorphism $$\sigma: G \to G/H \cong \{ \pm 1 \}$$, which I call the "alternating rep" by analogy with the symmetric group.

$$(\chi_{Ind(Res(U))}, \chi_U) =2$$, then, if and only if $$U \otimes \text{ alternating} \cong U$$. This then comes down to showing that:

$$\sum_{g \not \in H} |\chi(g)|^2 - \sum_{h \in H} |\chi(g)|^2 =0$$, since this is $$|G| \cdot (\chi_{U}, \chi_{U \otimes \text{alternating}})$$. Somehow, I was hoping that the assumption that whenever two elements of $$H$$ are conjugate in $$G$$, they are conjugate in $$H$$ would let me do something with the sums $$\sum_{h \in D_i} h \in \mathbb{C} [G]$$, since this assumption implies that this is an element of the center, and then Schur's Lemma tells me it acts by multiplication by constant.

But, I'm having trouble using this kind of reasoning!

What else can I do? How do you do this? Does my approach seem at all correct? Could you set me on the right path?