# Do field automorphisms of a character imply outer automorphisms of the group?

Apologies for the imprecise wording of the title.

In studying the basic representation theory of finite groups, I've been struck by a pair of phenomena present in every example I've worked with but not mentioned in any texts I've studied. I am wondering if either or both of these phenomena is true in general, or perhaps true under some assumptions that are usually satisfied for small groups.

Phenomenon #1: We are given a finite group $G$ and a representation $\rho:G\rightarrow GL(V)$ on a complex vector space $V$, with character $\chi$. Suppose $\chi(a)$ is an algebraic irrational, say $\alpha$, for some conjugacy class $a$ of $G$. Then for any conjugate $\alpha'$ of $\alpha$ (i.e. another root of $\alpha$'s minimal polynomial over $\mathbb{Q}$), there exists a conjugacy class $a'$ of $G$, of the same size as $a$, such that $\chi(a')=\alpha'$.

Phenomenon #2: Furthermore, there is an outer automorphism of $G$ that carries $a$ to $a'$.

Are these patterns at all general? If so, what assumptions are needed for them to hold? If not, what are some counterexamples?

Thanks so much!

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Have you tried looking at groups $G$ with $\mathrm{Out}(G)$ trivial, such as $S_n$? – Alex Becker Jan 6 '13 at 2:37
@Alex But all the characters of $S_n$ are rational. – Ted Jan 6 '13 at 2:44
@Ted Yes, I just looked that up. But that's the only class of groups with no outer automorphisms I can think of. – Alex Becker Jan 6 '13 at 2:45
The outer automorphism group of any nonabelian simple group is complete - i.e. it has trivial outer automorphism group. – Derek Holt Jan 6 '13 at 14:48
@DerekHolt - why is that? – Ben Blum-Smith Jan 6 '13 at 15:09

#1 is true. If $a$ has order $n$, then by conjugation, we may assume that $\rho(a)$ is diagonal with $n$th roots of unity on the diagonal. Thus $\alpha = \chi(a)$ is a sum of $n$th roots of unity. If $\alpha'$ is a conjugate of $\alpha$, then $\sigma(\alpha) = \alpha'$ for some $\sigma \in$ Gal($\mathbb{Q}(\mu)/\mathbb{Q}$) (where $\mu$ is a primitive $n$th root of unity). We must have $\sigma(\mu) = \mu^k$ for some $k$ prime to $n$, in which case $\sigma$ sends all $n$th roots of unity to their $k$th powers. Then $\rho(a^k)$ is also diagonal with all entries from $\rho(a)$ raised to the $k$th power. Hence $\chi(a^k) = \alpha'$.
The conjugacy classes of $a$ and $a^k$ have the same size because the $k$th power map and the $k^{-1}$ (mod $n$) power map are inverse maps between the two conjugacy classes.
Take $G=PGL(2,7)$, which has no outer automorphisms and has two non-rational characters. – Mariano Suárez-Alvarez Jan 6 '13 at 3:24