Any 1-dimensional character $\otimes$ irreducible character is irreducible I would like to prove that any 1-dimensional character $\otimes$ irreducible character is again irreducible. (I think this is character but there is chance I have misinterpreted my notes and it may be representation).
If $\chi_2$ is the irreducible character then we know that $\langle \chi_2,\chi_2 \rangle=1$. Let $\chi_1$ be the 1-dimensional character and so we know that $\chi_1=(1,\dots)$
I cannot see where to go from here
 A: Hint : if $\chi$ is irreducible and $1$-dimensional, then $\overline{\chi(g)} = \chi(g)^{-1}$ for every $g \in G$.
Now can you compute $\langle \chi_1 \otimes \chi, \chi_1 \otimes \chi \rangle$ ?
Proof of this fact : 
Let $(\rho, V)$ be a representation of a group $G$, with finite degree. The character of $g$ is the trace of the endomorphism $\rho(g)$. When a representation is $1$-dimensional, $\rho$ is equal to its character (the "trace" of a complex number is this complex number). When the group $G$ is finite with cardinal $n$, then $g^n = 1$, so $\rho(g)^n = 1$ and therefore, $\rho(g)$ is a root of unity, hence the equality $\rho(g^{-1}) = \rho(g)^{-1} = \overline{\rho(g)}$.
The preceding is not true in general, for example if the group $G$ is not finite : a representation (or character) of $\mathbb{Z}$ is given by $\rho : n \to a^n$, for any $a \neq 0$, and we have for example $\rho(-2) = a^{-2} \neq \overline{a^2}$ for most choices of $a$. 
Also, it need not be true for representations of degree $d$ higher than $1$ : for example, the character of the null element is always $d$. But, $\chi(e)^{-1} = 1/d$, which is not $\bar{d} = d$.
Answer to the question. Using the previous fact, we'll compute the "inner product" :
$$\langle \chi_1 \otimes \chi, \chi_1 \otimes \chi \rangle = \frac{1}{|G|} \sum_{g \in G} \chi_1(g)\chi(g)\overline{\chi_1(g)}\overline{\chi(g)}$$
$$= \frac{1}{|G|} \sum_{g \in G} \chi_1(g)\chi(g)\overline{\chi_1(g)}\chi(g)^{-1} = \frac{1}{|G|} \sum_{g \in G} \chi_1(g)\overline{\chi_1(g)}$$
$$= \langle \chi_1, \chi_1 \rangle $$
this latter quantity is equal to $1$ because the character $\chi_1$ is irreducible, therefore $\langle \chi_1 \otimes \chi, \chi_1 \otimes \chi \rangle = 1$, proving the irreducibility of the character $\chi_1 \otimes \chi$.
