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Homesecond proof of Wedderburn's theorem
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second proof of Wedderburn’s theorem
We can prove Wedderburn’s theorem,without using Zsigmondy’s theorem on the conjugacy class formula of the first proof;
let $G_{n}$ set of nth roots of unity and $P_{n}$ set of nth primitive
roots of unity and $\Phi_{d}(q)$ the dth cyclotomic polynomial.
It results

$\Phi_{n}(q)=\prod_{{\xi\in P_{n}}}(q\xi)$

$p(q)=q^{n}1=\prod_{{\xi\in G_{n}}}(q\xi)=\prod_{{d\mid n}}\Phi_{d}(q)$

$\Phi_{n}(q)\in\mathbb{Z}[q]\;$, it has multiplicative identity and $\Phi_{n}(q)\mid q^{n}1$

$\Phi_{n}(q)\mid\frac{q^{n}1}{q^{d}1}\;$with $d\mid n,d<n$
by conjugacy class formula, we have:
$q^{n}1=q1+\sum_{x}\frac{q^{n}1}{q^{{n_{x}}}1}$ 
by last two previous properties, it results:
$\Phi_{n}(q)\mid q^{n}1\;,\;\Phi_{n}(q)\mid\frac{q^{n}1}{q^{{n_{x}}}1}% \Rightarrow\Phi_{n}(q)\mid q1$ 
because $\Phi_{n}(q)$
divides the left and each addend of $\sum_{x}\frac{q^{n}1}{q^{{n_{x}}}1}$
of the right member of the conjugacy class formula.
By third property
$q>1\;,\;\Phi_{n}(x)\in\mathbb{Z}[x]\Rightarrow\Phi_{n}(q)\in\mathbb{Z}% \Rightarrow\Phi_{n}(q)\mid q1\Rightarrow\Phi_{n}(q)\leqslant q1$ 
If, for $n>1$,we have $\Phi_{n}(q)>q1$, then $n=1$ and the theorem is proved.
We know that
$\Phi_{n}(q)=\prod_{{\xi\in P_{n}}}q\xi\;,\;with\;q\xi\in\mathbb{C}$ 
by the triangle inequality in $\mathbb{C}$
$q\xi\geqslantq\xi=q1$ 
as $\xi$ is a primitive root of unity, besides
$q\xi=q1\Leftrightarrow\xi=1$ 
but
$n>1\Rightarrow\xi\neq 1$ 
therefore, we have
$q\xi>q1=q1\Rightarrow\Phi_{n}(q)>q1$ 
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