# Show that character $\chi_ϕ(g) ∈ \Bbb Z$ for any representation $ϕ$ of $G$.

Let $G$ be a finite group, $g ∈ G$ be an element of order $n$. Suppose that, for each $1 ≤ i ≤ n$ with $(i, n) = 1$, $g$ is conjugate to $g^i$ . Then show that $\chi_ϕ(g) ∈ \Bbb Z$ for any representation $ϕ$ of $G$.

Here is my try:

As $G$ is finite so $\phi(g)=diag(a_1,...,a_n)$ where $a_i$ are any $n$th root of unity. Then $\chi_ϕ(g)=a_1+....+a_n$.

Now $\chi_ϕ(g) \in \Bbb Q(w)$ where $w=e^\frac{2\pi i}{n}$

Then $\sum_{\sigma \in Gal(\Bbb Q(w)/\Bbb Q)} \sigma(\chi_ϕ(g)) \in \Bbb Q$ Now $\sum_{\sigma \in Gal(\Bbb Q(w)/\Bbb Q)} \sigma(\chi_ϕ(g))=\sum_{\sigma \in Gal(\Bbb Q(w)/\Bbb Q)}(\sigma(a_1)+...+\sigma(a_n))=\sum_{j=1}^n\sum_{\{i: (i,n)=1, 1 \leq i\leq n\}}a_j^i \in \Bbb Z$.

[As $\sum_{\{i: (i,n)=1, 1 \leq i\leq n\}}a_j^i \in \Bbb Z$].

From here how can we prove $\chi_ϕ(g) ∈ \Bbb Z$?

• A closely related thread. Possibly even a duplicate, though you probably weren't around at that time, and could not possibly remember. So +1s to you both for keeping interesting questions on the front page :-) – Jyrki Lahtonen Mar 17 '17 at 20:15

You have identified that each $\chi(g^i)$ is in the field $\mathbb Q(\omega)$, and you have shown that $\sum_{(i,n) = 1} \chi(g^i)$ is invariant under the action of any automorphism $\sigma \in Gal(\mathbb Q(\omega) : \mathbb Q)$. You then deduced that $$\sum_{(i,n) = 1} \chi(g^i) \in \mathbb Q.$$ This is most of the hard work!
Since $g$ is conjugate to $g^i$ when $(i,n) = 1$, you know that $\chi(g) = \chi(g^i)$ when $(i,n) = 1$. Therefore, $\chi(g^i) \in \mathbb Q$ for any $i$ such that $(i,n) = 1$.
But characters of representations over $\mathbb C$ are always algebraic integers! (See here.)
The only algebraic integers in $\mathbb Q$ are the (genuine) integers! So you can conclude that $\chi(g^i) \in \mathbb Z$ for any $i$ such that $(i,n) = 1$.