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Given a finite group $G$, and a non-identity representative $g$ in a conjugacy class of prime order $p$, I'm trying to show that some nontrivial irreducible character of $G$ must have $\chi(g) \neq 0$ and $\chi(1) \neq 0 \text{ mod } p$.

The suggestion was to show that the nonexistence of such a character would imply that $1/p$ is an algebraic integer.

So far, I've used the fact that $\chi(g) \cdot \frac{p}{\chi(1)} \in \overline{\mathbb{Z}}$ for any character. With some manipulation (and closure of algebraic integers over sums and products), I can get this into a form that allows me to sum over all irreducible characters $$p \sum \chi_i(g)^2 \in \overline{\mathbb{Z}}.$$ My hope was to then use the orthogonality relations on the columns of a character table, but doing so leaves a pesky factor of the order of the group, i.e. $|G|/p \in \overline{\mathbb{Z}}$ instead of $1/p$.

Am I going about this the wrong way? Is there something else I am overlooking?

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The column orthogonality relations for character tables give that $\sum_i \chi_i(g) \overline{\chi_i(1)} = 0$. But if every nontrivial character has either $\chi_i(g) = 0$ or $\chi_i(1) \equiv 0 \text{ mod } p$, then we have that $$\sum_{\text{nontrivial characters}} \chi_i(g) \overline{\chi_i(1)} \equiv 0 \text{ mod } p,$$ and then $\sum_i \chi_i(g) \overline{\chi_i(1)} \equiv 1 \text{ mod } p$ due to the trivial character - a clear contradiction.

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  • $\begingroup$ Sorry to nitpick, but mod p doesn't really make sense we're not working in $ $\endgroup$ – user149792 Nov 6 '14 at 23:32
  • $\begingroup$ You're right - so this answer doesn't really make sense. $\endgroup$ – Empiromancer Nov 17 '14 at 5:08
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You were almost there: let $p$ be a prime, and assume that $g \neq 1$ and $p \mid \chi(1)$ or $\chi(g)=0$ for every non-trivial $\chi \in Irr(G)$. (This implies that $G$ does not have any non-trivial linear characters!) Put $Irr(G)^{\#}=Irr(G)-\{1_G\}$. Column orthogonality tells us $$\sum_{\chi \in Irr(G)}\chi(g)\chi(1)=0$$, since $g$ is not conjugate to $1$. But $$0=\sum_{\chi \in Irr(G)}\chi(g)\chi(1)=1+\sum_{\chi \in Irr(G)^{\#}}\chi(g)\chi(1)=\\1+\sum_{\chi \in Irr(G)^{\#}, \chi(g)=0}\chi(g)\chi(1)+\sum_{\chi \in Irr(G)^{\#}, \chi(g) \neq 0}\chi(g)\chi(1)\\=1+p\sum_{\chi \in Irr(G)^{\#}, \chi(g) \neq 0}\chi(g)\frac{\chi(1)}{p}.$$ Hence $$\frac{-1}{p}=\sum_{\chi \in Irr(G)^{\#}, \chi(g) \neq 0}\chi(g)\frac{\chi(1)}{p}.$$ Now the right-hand side is an element of $\mathbb{A}$, the algebraic integers, and the left hand side is a rational. Since $\mathbb{Q} \cap \mathbb{A}=\mathbb{Z}$, we now have a contradiction.

Remark So we have proved that for any prime $p$ there must be a non-trivial character $\chi \in Irr(G)$ and $g \neq 1$, such that $p \nmid \chi(1)$ and $\chi(g) \neq0$. This statement is trivially true when $G$ has non-trivial linear characters. But becomes interesting when $G=G'$, that is, $G$ is perfect. The condition above on the size of the conjugacy class is not needed. In case, with the $g$ above, $|Cl_G(g)|=p$, then $|\chi(g)|=\chi(1)$. This follows from a theorem of Burnside (see for example Theorem (3.8), in M.I. Isaacs, Character Theory of Finite Groups)

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