Let $G_1$ and $G_2$ be finite groups such that $G_1$ is a subgroup of $G_2$. Let $V$ be a representation of $G_1$ (over some field; I am not assuming that the characteristic of the field is $0$ or that $V$ is finite dimensional). There is a natural $G_1$-equivariant map $\phi : V \rightarrow \text{Ind}_{G_1}^{G_2} V$. Let $W$ be a $G_2$-subrepresentation of $\text{Ind}_{G_1}^{G_2} V$ such that $W \neq \{0\}$.
Question : How do you show that $\phi(V) \cap W \neq \{0\}$?
This is false. Let $G_1$ be the trivial group, $G_2 = \mathbb{Z}/2$, and let $V = \mathbb{C}$ the trivial representation. Call the induced representation $W$; we have $W = V \oplus \sigma V$ where $\sigma$ is the nontrivial element of $G_2$. For any nonzero $c\in V$, the span of $c + \sigma c$ is a $G_2$-subrepresentation that has trivial intersection with $V$.