# Prove that every non-abelian group of order $8$ has a $2$-dimensional irreducible character all of whose values are integers.

Prove that every non-abelian group of order $8$ has a $2$-dimensional irreducible character all of whose values are integers.

I get that if $G$ is non abelian it must have an irreducible representation $>1$ and this implies that $$8=1^2+1^2+1^2+1^2+2^2$$ is the only possibility. Hence unique $2$-dimensional irreducible character,

I cannot see how to show that all its values will be integers. I feel this question may need some Galois theory that I am not very familiar with.

EDIT: I have the following hints.

$1)$ If $\chi$ is a character then so is $\sigma(\chi)$ for every automorphism $\sigma$ of $\mathbb{C}$.

$2)$ If $\alpha \in \mathbb{C}$ is fixed by every automorphism of $\mathbb{C}$ then $\alpha \in \mathbb{Q}$.

• This is definitely the wrong way of solving the problem, but there are only two nonabelian groups or order $8$; you can find such a character explicitly. – anomaly Apr 27 '15 at 18:07

Consider complex conjugation as an automorphism of $\mathbb C$. The complex conjugate of a character is also a character. You only have one character of degree $2$, so that implies something about the values.

Do you know that the values of a character are algebraic integers? Here, they are the sums of eigenvalues of matrices which satisfy $M^4=I$ (the orders of elements in a non-abelian group of order $8$ all divide $4$), and therefore satisfy $\lambda^4=1$. So if you don't know a general theorem about algebraic integers, you can simply list the possible sums of two eigenvalues, and see what the possibilities are.

• @JyrkiLahtonen I knew that - thanks for reminding me. I was thinking of automorphisms which fix $\mathbb R$ - but this is a technicality the answer can do without (especially as the question is working over $\mathbb Q$), so I've edited it out altogether per your comment. – Mark Bennet Apr 28 '15 at 12:06
• @MarkBennet Your first paragraph means that the values will all be real. I know from a theorem that the character have to be algebraic integers but this doesnt imply they are integers. Or does it? – Permian Apr 28 '15 at 12:11
• @sandstone I was picking up from the other answer a possible factor of $2$ which did not immediately exclude half integer values - any rational values are integers. But here you can just compute the cases. – Mark Bennet Apr 28 '15 at 12:19
• I actually used the fact that character values are algebraic integers to convince myself that the factor of $2$ wasn't a problem. This answer is a much more elegant way to look at it, because it proves in one sentence that a character which is unique of given degree always takes integer values. – Slade Apr 29 '15 at 19:05
• I have a theorem that says $\text{algebraic integers} \cap \mathbb{Q}=\mathbb{Z}$. Does this work now? – Permian Apr 30 '15 at 8:48

If $\chi_1, \ldots , \chi_4$ are the degree $1$ characters, and $\chi_5$ is the degree $2$ character, then what is $\chi_1 + \chi_2 + \chi_3+ \chi_4 + 2\chi_5$?

• I dont see how youve got this or where it goes – Permian Apr 27 '15 at 19:19
• @sandstone, the given sum is (by general theorey) the character of the regular representation, and therefore its value on an element of $g$ is $0$ (if $g \neq 1$) or $8$ (if $g=1$). This lets you write the values of the two-dimensional character in terms of the one-dimensional characters. – Stephen Apr 27 '15 at 20:07
• @Stephen I understand your explanation but I dont have explicitly $\chi_1,\chi_2,\dots,\chi_5$ so how could I evaluate it – Permian Apr 28 '15 at 11:13
• @sandstone The $1$-dimensional characters are the characters of the abelianization, which happens to be the quotient by the center when the group has order $8$. – Tobias Kildetoft Apr 28 '15 at 12:07
• @sandstone The quotient of a non-abelian group by its center is never cyclic (this is a quite well-known result, and it is not too hard to prove as an exercise). Thus, in this case it must be a group where all elements have order dividing $2$, so the only possible values of the irreducible characters are $1$ and $-1$. – Tobias Kildetoft Apr 28 '15 at 12:17