Navarro Character Theory Exercises 4.5 Recently, I am studying the book Navarro - Character Theory and the McKay Conjecture. I am trying to solve the following exercise:
(Exercise 4.5)
Let $G$ be a finite group, suppose that $\alpha,\beta\in{\rm Irr}(G)$, and let $p$ be a prime not dividing $|G|$. If $\alpha^p=\beta^p$, then prove that $\alpha=\beta$.
If $p$ is odd, the exercise is trivial. But I can't show $\alpha=\beta$ when $p=2$.
Could you give me some hints then.
Thanks a lot in advance.
 A: Let $G$ be a finite group of odd order and let $\alpha, \beta \in Irr(G)$ be irreducible complex characters, such that $\alpha^2=\beta^2$. 
First we note the following: if $g,h \in G$ with $g^2=h^2$, then $g=h$. For $|G|=2a+1$ for some positive integer $a$, whence (using Lagrange's Theorem) $g=g^1=g^{|G|-2a}=g^{|G|} \cdot (g^2)^{-a}= 1 \cdot (h^2)^{-a}=h^{|G|-2a}=h$. 
Still another thing to note in odd ordered groups: other than the identity class, the conjugacy classes come in pairs: if $g \in G -\{1\}$, then  $g \not\sim g^{-1}$. For if $x^{-1}gx=g^{-1}$, then $gx=xg^{-1}$, so $(xg)^2=x(gx)g=x(xg^{-1})g=x^2$ and by the note above it follows that $xg=x$, hence $g=1$, a contradiction. It follows that $Cl_G(g)$ and $Cl_G(g^{-1})$ are different (disjoint, if you will) conjugacy classes (by the way, of the same cardinality, but we do not use this). It is easy to see that for a fixed $g \in G$, the class function $|\alpha|^2$ takes the same value on $Cl_G(g)$ and $Cl_G(g^{-1})$. 
Now back to our irreducible characters $\alpha$ and $\beta$. If $g \in G$, $\alpha(g)^2=\beta(g)^2$, then $(\alpha(g)+\beta(g)(\alpha(g)-\beta(g)=0$ and working over the complex numbers, we have $\alpha(g)=\pm \beta(g)$. Observe that $\alpha(g)=0$ if and only if $\beta(g)=0$. Hence we can write $G$ as a disjoint union of normal sets (e.g. invariant under $G$-conjugation):
$$G=N \cup S \cup T$$
where $N=\{g \in G: \alpha(g)=0=\beta(g)\}$ (this set could be empty in case $\alpha$ and hence $\beta$ are linear - the problem is trivially solved then), $S=\{g \in G-N: \alpha(g)=\beta(g)\}$, $T=\{g \in G-N: \alpha(g)=-\beta(g)\}$. 
$S$ consists of $\{1\}$ and pairs of conjugacy classes $Cl_G(g)$ and $Cl_G(g^{-1})$ with $g \in S-\{1\}$. Let the set $U \subset S$ denote half of these pairs. Let $V \subset T$ denote the similar set of conjugacy classes contained in $T$. Note that $1 \in S$. To reach a contradiction, assume $\alpha \neq \beta$. Then the Frobenius inner product $[\alpha,\beta]=0$. It follows that
$$0=\alpha(1)\beta(1)+ \sum_{g \in S-\{1\}}\alpha(g)\overline{\beta(g)}+\sum_{g \in T}\alpha(g)\overline{\beta(g)}$$
implying $$-\alpha(1)\beta(1)=\sum_{g \in S-\{1\}}\alpha(g)\overline{\beta(g)}+\sum_{g \in T}\alpha(g)\overline{\beta(g)} \\
=\sum_{g \in S-\{1\}}|\alpha(g)|^2-\sum_{g \in T}|\alpha(g)|^2 \\
= \sum_{\text{disjoint pairs } \{g,g^{-1}\} \subseteq S-\{1\}}|\alpha(g)|^2+|\alpha(g^{-1})|^2-\sum_{\text{disjoint pairs } \{g,g^{-1}\} \subseteq T}|\alpha(g)|^2+|\alpha(g^{-1})|^2 \\
= \sum_{U}2|\alpha(g)|^2 -\sum_{V}2|\alpha(g)|^2.$$
From the last equation it follows that $\frac{-\alpha(1)\beta(1)}{2}$ is an algebraic integer. However, both $\alpha(1)$ and $\beta(1)$ divide the order $|G|$ and so are both odd numbers as is their product. We now have reached a contradiction, since half an odd integer is not an algebraic integer. $\square$
