Show $U \otimes V$ is an irreducible G-module Let $G$ is some group and $U$ is an irriducible $G$-module over the complex numbers. Now if $V$ is a $G$-module of dimension 1, I would like to prove $U \otimes V$ is an irriducible $G$-module. 
My knowledge of tensor products is lacking..
 A: Another, definitively less elegant, method to solving this is to recall that a representation $\rho:G\to\text{GL}_n(\mathbb{C})$ is irreducible if and only if $\langle\chi_\rho,\chi_\rho\rangle=1$, and that if $\rho,\psi:G\to\text{GL}_n(\mathbb{C})$ are a pair of representations then 
$$\chi_{\rho\otimes\psi}=\chi_\rho\chi_\psi$$
With this in mind, let $\rho:G\to\text{GL}_n(\mathbb{C})$ be an irreducible representation and $\psi:G\to S^1$ a character. Then,
$$\begin{aligned}\langle\chi_{\rho\otimes\psi},\chi_{\rho\otimes\psi}\rangle &= \frac{1}{|G|}\sum_{g\in g}|\chi_{\rho\times\psi}(g)|^2\\ &= \frac{1}{|G|}\sum_{g\in G} |\chi_\rho(g)|^2|\chi_\psi(g)|^2\\ &= \frac{1}{|G|}\sum_{g\in G}|\chi_\rho(g)|^2\\&=1\end{aligned}$$
where we used the fact that $\chi_\psi(g)=\psi(g)\in S^1=\{z:|z|=1\}$ for all $g$, and the last step was achieved using the fact that $\rho$ was irreducible.
This is much less elegant, and much more difficult than is needed, but it's always useful to remember this brute-force method of attacking irreducibility over $\mathbb{C}$.
A: Suppose that $\{u_1, u_2, \dots \}$ is a basis for the irreducible $G$-module $U$.  Since $V$ is $1$-dimensional, $V = \mathbb{C}v$ for some nonzero $v \in V$.
Then, a basis for $U \otimes V$ is $\{u_1 \otimes v, u_2 \otimes v, \dots \}$.  Any $G$-submodule of $U \otimes V$ will be of the form $W \otimes V$, where $W \le U$.  Why?  How does $G$ act in the tensor product?
As $U$ is irreducible, either $W = 0$ or $W = U$.  As a consequence, either $W \otimes V = 0 \otimes V = 0$ or $W \otimes V = U \otimes V$.  Therefore $U \otimes V$ is irreducible.
A: If $V$ has dimension $1$ then it is invertible in the sense that there's another module $V'$ such that $V \otimes V'$ is isomorphic to the trivial representation.  (A good exercise is to work out what this $V'$ is.)  If you have a submodule $W \subset U \otimes V$ then $W \otimes V' \subset U$, so any submodule of $U \otimes V$ gives a submodule of $U$ with the same dimension.
