I was writing this question, and I came up with an answer, so I thought I would answer it myself:
In considering representations of $S_n$, among others, we have the "sign representation", that is the one-dimensional representation
$$
\rho_{\Sigma}:S_n \to \mathbb{C}^{\times}: \tau \mapsto \text{sgn}(\tau).
$$
When we are finding irreducible representations of $S_n$, sometimes the argument in books I read proceeds as follows: we find an irreducible representation $\rho$ on $V$. We note that the character of $\rho \otimes \rho_{\Sigma}$ is distinct from that of $\rho$. Then we conclude that $\rho \otimes \rho_{\Sigma}$ must be another irreducible representation.
The character table tells us that $\rho \otimes \rho_{\Sigma}$ is different from $\rho$, but how do we know that it is irreducible?
Best Answer
You don't need any character theory to do this. Let $V$ be an irreducible representation of any group $G$ (the group is not necessarily finite and $V$ is not necessarily finite-dimensional) and let $L$ be a $1$-dimensional representation. I claim that $V \otimes L$ is still irreducible. The reason is that tensoring with $L$ is invertible: the natural map $L^{\ast} \otimes L \to 1$ (where $1$ is the trivial representation) is an isomorphism, so
$$(V \otimes L) \otimes L^{\ast} \cong V.$$
Consequently, if $W$ is a proper nonzero submodule of $V \otimes L$, then $W \otimes L^{\ast}$ is a proper nonzero submodule of $V$. More abstractly, tensoring with $L$ is an automorphism of the category of representations of $G$, and automorphisms of categories preserve categorical properties of their objects like irreducibility.