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Let $C$ be a conjugacy class of the finite group $G$. Say that $C$ is rational if for each character $\chi: G \rightarrow \mathbb C$ of $G$, for each $c\in C$, we have $\chi(\sigma) \in \mathbb Q$. I am trying to show that $C$ is rational if and only if whenever $c\in C$ and $n$ is relatively prime to $|c|$, we have $c^n \in C$. Any suggestions?

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Let $\mathbb{Q}(\chi)$ be a finite normal extension of $\mathbb{Q}$ containing the coefficients of $\chi'(g)$ for all $g \in G$, where $\chi'$ is the representation associated with $\chi$. Let $m=|G|$ and $\epsilon$ be a primitive $m$th root of unity. Let $E$ be a finite normal extension of $\mathbb{Q}$ containing $\mathbb{Q}(\chi)$ and $\mathbb{Q}(\epsilon)$. Then there is an injective homomorphism $\psi:\mathbb{Z}_m^*\rightarrow G(E,\mathbb{Q})$ defined by $\psi(a)=\sigma_a$, where $\sigma_a(\epsilon)=\epsilon^a$. Now define $\chi_a' = \tau_a\circ \chi'$ (where $\tau_a$ is the automorphism of $GL_m(E)$ induced by applying $\sigma_a$ to the coefficients of each matrix element) and let $\chi_a$ be its character. Then $\chi_a(g)=\chi(g^a)$ (why?).

In one direction, $c$ is conjugate to $c^n$ for $\left(n,m\right)=1$, so $\chi(c)=\chi(c^n)$. Since $\chi(c^n)=\chi_n(c)$, we have that $\chi(c)$ is fixed under the action of $\sigma_n$. If this is true for every $n$ relatively prime to $m$, then $\chi(c)$ is fixed under the image of $G(\mathbb{Q}(\epsilon),\mathbb{Q})$. Thus $\chi(c)$ is a member of the fixed field of $G(\mathbb{Q}(\epsilon),\mathbb{Q})$, otherwise known as $\mathbb{Q}$.

On the other hand, suppose $C$ is rational and for some $\chi$ there's an $c\in C$ and $n$ relatively prime to $m$ for which $\chi(c)\not= \chi(c^n)$, then $\chi(c)$ is not fixed under $\sigma_n$, so it is not a member of the fixed field of $G(\mathbb{Q}(\epsilon),\mathbb{Q})$, a contradiction. So $\chi(c)=\chi(c^n)$ for every character $\chi$ of $G$, whence $$\sum_i \chi_i(c)\overline{\chi_i(c^n)}=\sum_i\left|\chi_i(c)\right|^2= |C_G(c)|$$ where the summation runs over all irreducible characters, which implies that $c$ is conjugate to $c^n$.

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There are a few lines here which I am unsure about. (i) I believe that the domain of $\psi$ should be $\mathbb Z_m^\times$ (for example, $a = 0\in \mathbb Z_m$ does not define an automorphism). (ii) How do we know $\chi_a(g) = \chi(g^a)$ for all $g$? I could not find a way to prove this. (iii) You establish that $\chi(c)$ belongs to the fixed field of $G(\mathbb Q(\epsilon) , \mathbb Q)$. This is a subfield of $E$, but if $E$ properly contains $\mathbb Q(\epsilon)$, this fixed field could be large than $\mathbb Q$. –  user15464 Oct 22 '12 at 3:49
    
Yes, you are right, I meant $\mathbb{Z}_m^*$. For $\chi_a(g)=\chi(g^a)$, we have that $\chi'(g)$ is diagonalizable so that its diagonal entries are $k$th roots of unity, for $k=|g|$. Thus $\chi(g)$ is a sum of $k$th roots of unity, so we can write $\chi(g)=\epsilon_1+\ldots+ \epsilon_m$. Since $\chi'$ is a homomorphism, $\chi'(g^a)=\chi'(g)^a$, which is then similar to the diagonal matrix with diagonal entries $\epsilon_1^a,\ldots,\epsilon_m^a$, whence $\chi(g)^a=\chi(g^a)=\epsilon_1^a+\ldots +\epsilon_m^a$. –  Alexander Gruber Oct 22 '12 at 4:07
    
The action of $\tau_a$ on the $\chi'(g)$ raises all of its $m$th (and thus $k$th) roots of unity to $a$ where $(a,m)=(a,k)=1$. So $\chi_a(G)=\epsilon_1^a+\ldots + \epsilon_m^a$, which is exactly $\chi(g^a)$. –  Alexander Gruber Oct 22 '12 at 4:07
    
Ok, I got it. Thanks. I was just confused about the definition of $\tau_a$. This turned out a bit trickier than I expected, but certainly a cool proof. –  user15464 Oct 22 '12 at 4:14
    
Doesn't the last paragraph only show that $\chi(c)=\chi(c^n)$, not that $c$ and $c^n$ are actually conjugate? That seems to me to be the harder part of the desired proof. What you've written only relates to the similarity of matrices, not to the structure of the group and whether it actually contains elements that induce the required similarity. –  joriki Oct 22 '12 at 5:10
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