3
$\begingroup$

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 $\alpha \in \mathbb{Q}$.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – anomaly Apr 27 '15 at 18:07
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – Mark Bennet Apr 28 '15 at 12:06
  • $\begingroup$ @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? $\endgroup$ – Permian Apr 28 '15 at 12:11
  • $\begingroup$ @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. $\endgroup$ – Mark Bennet Apr 28 '15 at 12:19
  • $\begingroup$ 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. $\endgroup$ – Slade Apr 29 '15 at 19:05
  • $\begingroup$ I have a theorem that says $\text{algebraic integers} \cap \mathbb{Q}=\mathbb{Z}$. Does this work now? $\endgroup$ – Permian Apr 30 '15 at 8:48
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I dont see how youve got this or where it goes $\endgroup$ – Permian Apr 27 '15 at 19:19
  • $\begingroup$ @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. $\endgroup$ – Stephen Apr 27 '15 at 20:07
  • $\begingroup$ @Stephen I understand your explanation but I dont have explicitly $\chi_1,\chi_2,\dots,\chi_5$ so how could I evaluate it $\endgroup$ – Permian Apr 28 '15 at 11:13
  • 1
    $\begingroup$ @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 $8$. $\endgroup$ – Tobias Kildetoft Apr 28 '15 at 12:07
  • 1
    $\begingroup$ @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 $-1$. $\endgroup$ – Tobias Kildetoft Apr 28 '15 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.