Sorry to resurrect such an old thread, but we supply two proofs. The first proof is due to Sameer Kailasa.

**Problem 2.37, Fulton-Harris.** Show that if $V$ is a faithful representation of $G$, i.e., $\rho: G \to GL(V)$ is injective, then any irreducible representation of $G$ is contained in some tensor power $V^{\oplus n}$ of $V$.

Let $W$ be an irreducible representation of $G$, and set$$a_n = \langle \chi_W,\chi_{V^{\oplus n}}\rangle = \langle\chi_W,(\chi_V)^n\rangle.$$If we consider the generating function $f(t) = \sum_{n=1}^\infty a_nt^n$, we can evaluate it as$$f(t) = {1\over{|G|}}\sum_{n=1}^\infty \sum_{g\in G} \overline{\chi_W(g)}(\chi_V(g))^nt^n = {1\over{|G|}} \sum_{g \in G} \overline{\chi_W(g)} \sum_{n=1}^\infty (\chi_V(g)t)^n$$$$={1\over{|G|}} \sum_{g \in G}{{\overline{\chi_W(g)}\chi_V(g)t}\over{1 - \chi_V(g)t}}.$$Note that in this sum, the term where $g = e$ evaluates to $${{(\dim W \cdot \dim V)t}\over{1 - (\dim V)t}},$$which is nonzero. If no other term in the summation has denominator $1 - (\dim V)t$, then this term can not cancel, so $f(t)$ is a nontrivial rational function. We can then conclude that not all of the $a_n$ are $0$. Thus, to complete the proof, it suffices to show $\chi_V(g) = \dim V$ only for $g = e$.

Suppose $\chi_V(g) = \dim V = n$ for $g \neq e$. Also, say $G$ acts on $V$ via $\rho: G \to GL(V)$. There is $k$ such that $\rho(g)^k = I$. If $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $g$ we have$$\lambda_1^{ik} + \dots + \lambda_n^{ik} = n$$for $i = 0, 1, \dots$. Since $g^{k+1} = g$, we also see$$\lambda_1^{ik+1} + \dots + \lambda_n^{ik+1} = n.$$It follows that $$\lambda_1^{ik}(\lambda_1 - 1) + \dots + \lambda_n^{ik}(\lambda_n - 1) = 0,$$which implies for all polynomials in $\mathbb{C}[x]$, we have$$P(\lambda_1^k)(\lambda_1 - 1) + \dots + P(\lambda_n^k)(\lambda_n - 1) = 0.$$Choosing appropriate polynomials with roots at all but one of the eigenvalues, we see that all the eigenvalues must be $1$. Since $\rho(g)$ is diagonalizable, it follows $\rho(g) = I$. This contradicts the faithfulness of $V$.

**Problem 3.26, Etingof.** Let $G$ be a finite group, and $V$ a complex representation of $G$ which is faithful, i.e., the corresponding map $G \to GL(V)$ is injective. Show that any irreducible representation of $G$ occurs inside $S^nV$ (and hence inside $V^{\otimes n}$) for some $n$.

Let $n = |G|$.

## Step 1.

**There exists $u \in V^*$ whose stabilizer is $1$.**

For given $g \neq 1$, since $\rho_V:G \to GL(V)$ is injective, $\rho_V(g)^{-1} - I = \rho_V(g^{-1}) - I \neq 0$. Thus there exists $u \in V^*$ for which $(\rho_{V^*}(g) - I)u$ is not the zero transformation. (We make the observation that $((\rho_{V^*}(g) - I)u)(v) = u((\rho_V(g)^{-1} - I)v)$; just define $u$ so that it sends something in the range of $\rho_V(g)^{-1} - I$ to $1$.) Define$$U_g=\{u\in V^*\text{ }|\text{ }(\rho_{V^*}(g)-I)u= 0\};$$that is, $U_g$ is the kernel of the linear transformation $\rho_{V^*}(g) - I$ on $V^*$. Then when $g \neq 1$, $U_g$ is a proper subspace of $V^*$. Hence, the union $\bigcup_{g \in G,\,g \neq 1} U_g$ cannot be the entire space $V^*$. (See the following lemma.)

**Lemma.** *Let $W$ be a complex vector space and $W_1, \dots, W_m$ proper subspaces of $W$. Then$$W \neq \bigcup_{i=1}^m W_i.$$*

*Proof.* For each $i$, choose a vector $w_i \notin W_i$. Let $U = \text{span}(w_1, \dots, w_m)$. Note that $U \not\subseteq W_i$ for any $I$. Replacing $W_i$ with $W_i \cap U$ and $W$ with $U$ as necessary, we may assume that $W$ is finite-dimensional.

For each $i$, find a linear functional $f_i$ such that $\text{ker}(f_i) = W_i$. Choose a basis $e_1, \dots, e_k$ of $W$. Then$$f(x_1, \dots, x_k) := \prod_{i=1}^m f_i(x_1e_1 + \dots + x_ke_k)$$is a polynomial in the $x_1, \dots, x_k$ over an infinite field, so there exists $(x_1, \dots, x_k)$ such that $f(x_1, \dots, x_k) \neq 0$. This point is not in any of the $W_i$.$$\tag*{$\square$}$$Taking $u \in V^* - \bigcup_{g \in G} U_g$, we get that$$u \notin U_g \implies \rho_{V^*}(g)u \neq u$$for any $g \in G$, $g \neq 1$. In other words, $\rho_{V^*}u = u$ if and only if $g = 1$, and the stabilizer of $u$ is $1$.

## Step 2.

**Define a map $SV \to F(G, \mathbb{C})$.**

Define the map $\Phi: SV \to F(G, \mathbb{C})$ by sending $f \in SV$ to $f_u$ defined by $f_u(g) = f(gu)$. In other words, we define $\Phi$ as follows.

- First, define $\Phi_k: S^kV \to F(G, \mathbb{C})$ as the linear map induced by the symmetric $k$-linear map $\beta_k: V^k \to F(G, \mathbb{C})$ given by$$[\beta_k(v_1, \dots, v_k)](g) = \prod_{i=1}^k[(\rho_{V^*}(g)u)(v_i)] = \prod_{i=1}^k [i(\rho_V(g)^{-1}v_i)].$$Note that $\Phi_k$ is a homomorphism of representations since$$[\Phi_k(h(v_1\dots v_k))](g) = [\Phi((hv_1) \dots (h v_k))](g) = \prod_{i=1}^k[(gu)(gv_i)]$$$$= \prod_{i=1}^k [(h^{-1}gu)(v_i)] = [\Phi_k(v_1 \dots v_k)](g^{-1}g) = \{h[\Phi_k(v_1 \dots v_k)]\}(g).$$(For $k = 0$, the map is the map $\mathbb{C} \to F(G, \mathbb{C})$ sending a number to its constant function.)
- Define $\Phi: SV \to F(G, \mathbb{C})$ by$$\Phi = \bigoplus_{k=0}^\infty \Phi_k.$$

## Step 3.

**$\Phi$ is surjective; in fact, the map restricted to $\bigoplus_{i \le n-1} S^i V$ is surjective.**

It suffices to show the functions $1_h$ defined by$$1_h(g) = \begin{cases} 1 & \text{if }g = h \\ 0 & \text{if }g \neq h \end{cases}$$are in the image of $\Phi$, since they span $F(G, \mathbb{C})$. Given $h$, we will find a vector $f \in SV$ such that $\Phi(f) = k1_h$ for some $k \in \mathbb{C} - \{0\}$.

Let $K$ be the kernel of $u$; since $u$ is a nontrivial linear transformation $V \to \mathbb{C}$,$$\dim(K) = \dim(V) - \dim(\mathbb{C}) = n-1.$$For each $g \in G$, let$$V_g = gK = \rho_V(g)K.$$So $V_g$ is the subspace of vectors $v$ such that $g^{-1}v \in \text{ker}(u)$, i.e. $u(g^{-1}v) = 0$. We define $v_g$ for $g \neq h$; consider two cases.

- If $V_g \neq V_h$, define $v_g \in SV$ to be a vector in $V_g - V_h \subseteq V$. Note each $V_g$ has dimension $n-1$ since $g$ is invertible. ($V_g$, $V_h$ both have the same dimension, so neither is contained in the other.) Then$$[\Phi(v_g)](h) = u(h^{-1}v_g) \neq 0,\text{ }[\Phi(v_g)](g) = u(g^{-1}v_g) = 0.$$
- If $V_g = V_h$ and $g \neq h$, then let $v_g'$ be a vector in $V - V_g$. Then $u(g^{-1}v_g') = \lambda$ for some nonzero $\lambda$. Define $v_g \in SV$ to be the vector $v_g' - \lambda$. Note that$$[\Phi(v_g)](g) = u(g^{-1}v_g') - \lambda = 0.$$If $u(h^{-1}v_g') = \lambda$, then $gu = u(g^{-1}*)$ and $hu = u(h^{-1}*)$ would be identical linear transformations (they already agree on $V_g$ as they are identically zero there; $V_g + \text{span}(v_g') = V$), contradicting the fact that $U$ has stabilizer $1$. Hence, $u(h^{-1}v_g') \neq \lambda$ and$$[\Phi(v_g)](h) \neq 0.$$Now consider$$f = \prod_{g \neq h} v_g \in \bigoplus_{i \le n-1} S^i V.$$We have $[\Phi(f)](g) = 0$ for all $g \neq h$ since $[\Phi(v_g)](g) = 0$ for $g \neq h$. On the other hand, $[\Phi(v_g)](h) \neq 0$ for all $g \neq h$, so $[\Phi(f)](h) \neq 0$. Thus, $\Phi(f)$ is a multiple of $1_h$. Since this works for all $h$, $\Phi$ is surjective.

## Step 4.

**$W := \bigoplus_{1 \le n-1} S^i V$ contains every irreducible representation of $V$.**

Note that$$F(G, \mathbb{C}) \cong \text{Hom}_\mathbb{C}(\mathbb{C}G, \mathbb{C}) \cong (\mathbb{C}G)^* \cong \mathbb{C}G.$$The last isomorphism follows since $\chi_{\mathbb{C}G}$ is real, (as each $\rho_{\mathbb{C}G}(g)$ is real) and hence equal to its conjugate $\overline{\chi_{\mathbb{C}G}} = \chi_{(\mathbb{C}G)^*}$. Since $W$ maps surjective to $F(G, \mathbb{C}) \cong G\mathbb{C}$ via $\Phi$, $G\mathbb{C}$ must actually occur inside $W$. This is since$$\chi_W = \chi_{\text{ker}(\Phi)} + \chi_{W/\text{ker}(\Phi)} = \chi_{\text{ker}(\Phi)} + \chi_{\mathbb{C}G}.$$Since $G\mathbb{C}$ contains every irreducible representation, so does $\oplus_{i \le n-1} S^i V$. Thus, every irreducible representation occurs inside $S^i V$ for some $i \le n-1$.