# 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

• The character of the tensor product of the two corresponding reps is the product $\chi_1\chi_2$. Try computing $\langle \chi_1\chi_2,\chi_1\chi_2\rangle = |G|^{-1}\sum_{g\in G} \cdots$ -- you want to show this equals 1, because then $\chi_1\chi_2$ is irreducible. You'll need to use the fact that $\chi_1(g)$ is a root of 1, so $|\chi_1(g)|^2 = 1$ for any $g$. – Matthew Towers Apr 29 '15 at 9:39
• Why is $\chi_1(g)$ a root of $1$? – Permian May 29 '15 at 11:55
• The value of a one dimensional character on g is the eigenvalue by which g acts. If g has order n then g to the power n acts as the nth power of this eigenvalue and also as 1, so the eigenvalue is an nth root of 1. – Matthew Towers May 29 '15 at 12:45

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 $\chi_1 \otimes \chi$.

• Does $\overline{\chi(g)} = \chi(g)^{-1}$ not hold for any character? I also dont see how to use it as I get $\langle \chi_1\chi_2, \chi_1\chi_2 \rangle=\frac{1}{|G|}\sum_{g \in G}\chi_1(g)\overline{\chi_1(g)}\chi_2\overline{\chi_2(g)}$ – Permian May 29 '15 at 11:54
• 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, then $g^n = 1$, so $\rho(g)^n = 1$ and therefore, $\rho(g)$ is a root of unity, hence the equality $\rho(g^{-1}) = \rho(g)^{-1} = \overline{\rho(g)}$. – Tlön Uqbar Orbis Tertius May 29 '15 at 12:18
• 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 $\rho(-2) = a^{-2} \neq \overline{a^n}$ for most choices of $a$. – Tlön Uqbar Orbis Tertius May 29 '15 at 12:20
• ok, I'll add them. – Tlön Uqbar Orbis Tertius May 29 '15 at 12:21
• Great answer thanks – Permian May 29 '15 at 12:24