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In the question Faithful representations and tensor powers, several proofs demonstrate that for every faithful complex representation $V$ of a finite group $G$, every irreducible complex representation of $G$ is a subrepresentation of a tensor power $V^{\otimes n}$ of $V$ for some $n$.

The second proof in this answer, based on Etingof's hint in Introduction to representation theory, establishes a stronger result: every irreducible representation appears in some symmetric power $S^nV$ of the faithful representation $V$. Another simple proof, referenced in a comment, arrives at the same strengthened conclusion.

Other proofs, such as those employing simple analysis argument, field/ring extension or generating function — rely on the fact that the character of a tensor product $W_1\otimes_{\mathbb{C}} W_2$ satisfies $\chi_{W_1\otimes W_2}=\chi_{W_1}\cdot \chi_{W_2}$. However, character $\chi_{S^nV}$ of a symmetric power $S^nV$ does not adhere to a simple form like $\chi_{V^{\otimes n}}=(\chi_V)^n$. This discrepancy makes it challenging to improve these proofs to establish the strengthened result that every irreducible representation appears in some symmetric power.

This leads me to wonder:

can we prove directly that $\bigoplus_{d=0}^\infty V^{\otimes d}$ contains all irreducible complex representations of $G$ implies that $\bigoplus_{d=0}^\infty S^dV$ also contains all irreducible complex representations?

For example, is it possible to show that the injectivity of the map $\mathbb{C}[G]\rightarrow \prod_{d=0}^\infty\operatorname{End}_\mathbb{C}V^{\otimes d}$ implies the injectivity of the map $\mathbb{C}[G]\rightarrow \prod_{d=0}^\infty\operatorname{End}_\mathbb{C}S^dV$?

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  • $\begingroup$ for your first question, both statements are equivalent to $V$ being faithful, so the statements are equivalent. Maybe you're looking for some more direct argument that avoids faithfulness of $V$? not sure how to make that precise.. $\endgroup$ Commented Dec 7 at 19:28
  • $\begingroup$ @DanielCopeland Yes, I'm looking for some direct argument. Often, when people try to prove a result $A$, they transform $A$ into an equivalent but actually easier-to-prove result $B$. The proof of the equivalence is exactly what I’m looking for. $\endgroup$ Commented Dec 7 at 20:02

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