if $\rho: H \to \text{GL}_n({\bf C})$ is faithful then $\text{Ind}_H^G \rho$ is faithful How do I show that if $\rho: H \to \text{GL}_n(\mathbb{C})$ is faithful then $\text{Ind}_H^G \rho$ is faithful?
 A: As an alternative to Sameer's answer, you could argue as follows: If $V$ is the $H$-module corresponding to $\rho$, then $\text{Ind}_H^G \rho = {\mathbb C}[G]\otimes_{{\mathbb C}[H]} V$, which as a vector space decomposes as the direct sum of $g_i\otimes V$ for a choice $\{g_i\}_i$ of $H$-right-coset representatives in $G$, among which we choose $e$ as the representative for $H$ itself. Then, if $g\in G$ acts trivially on $\text{Ind}_H^G\rho$, on the one hand it preserves the summand $e\otimes V$ but on the other hand maps it into the summand $g_i\otimes V$ for which $g_i H = g H$. Hence $g=h\in H$, in which case $h$ acts on the summand $e\otimes V$ by $\text{id}\otimes\rho(h)$; since $\rho$ is faithful, we deduce $h=e$.
A: Let $\rho: H \to \text{GL}(W)$ be a faithful representation, and let $V = \text{Ind}^{G}_{H} \, W$, given by $\hat{\rho}: G \to \text{GL}(V)$. Suppose there is a non-identity $g\in G$ such that $g\in \text{ker} \, \hat{\rho}$. Choose $S$ a set of coset representatives. By Mackey's formula, we see that $$\chi_{V} (g) = [G:H] (\text{dim} \, W) = \sum_{gsH = sH} \chi_{W} (s^{-1} g s)$$ But by the triangle inequality $$ [G:H] (\text{dim} \, W) = \left\vert \sum_{gsH = sH} \chi_{W} (s^{-1} g s) \right\vert \le \sum_{gsH = sH} |\chi_{W} (s^{-1} g s)| \le k (\text{dim} \, W)$$ where $k$ is the number of cosets fixed by $G$. Clearly $k\le [G:H]$, so $g$ must fix every coset. In particular, this means $gH = H \implies g\in H$. Since elements of $H$ act on $W \subset V$ as $\rho$, it follows that $g\in \text{ker} \, \rho$, contradiction.
A: By using a stronger result, I will give a short and general answer.
Let $\chi$ be a the corresponding character of $H$ and $\chi^G$ be induced character of $G$.
Lemma: $Ker(\chi^G)=Core_G(Ker(\chi))=\bigcap (Ker\chi)^g$
Thus, If $Ker(\chi)=e$ then clearly $Ker(\chi^G)=e$. Hence, $\chi^G$ is faitful.
