A step in proof of Burnside's Theorem
I am reading the proof of Burnside's Theorem and it uses the following lemma in the online notes. page $70$ of - .
Let $s \in G-\{1\}$ and $p$ be a prime. Let the cardinality of conjugacy class of $s$ be $c(s)=p^\alpha$, for some $\alpha$.
Then $\exists$ $N\lhd G $ and $N\neq G$, such that,
If $ \phi : G\rightarrow G/N$ be the natural homomorphism, then $\phi (s) \in Z(G/N)$, where $Z(G/N)$ is the center of $G/N$.
Proof: Let $r_{G}$ be the character of regular representation of G. Therefore $$r_{G}(s)=\sum \chi(1)\chi(s)=0$$ where the sum is over all the irreducible characters of G.
Now, $$1+\sum_{\chi \neq 1} \chi(1)\chi(s)=0$$$\implies$$$1/p+\sum_{\chi \neq 1} (\chi(1)\chi(s))/p=0$$$\implies$$$-1/p=\sum_{\chi \neq 1} (\chi(1)\chi(s))/p$$
$\implies$ $-1/p$ isn't algebraic integer .
How is the last step justified? How do we know that $-1/p$ isn't an algebraic integer?
$\endgroup$ 61 Answer
$\begingroup$As per the comment by @anon
If $f(−1/p)=0$, $f(−1/p)=0$ for some monic integer-coefficient polynomial $f$ of degree $n$ then multiplying by $p^n$ and reducing mod $p$ yields $(−1)^n≡0$ then, $(−1)≡0$ or $1≡0$, a contradiction.
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