# 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 n-th roots of unity and $P_{n}$ set of n-th primitive roots of unity and $\Phi_{d}(q)$ the d-th 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

by conjugacy class formula, we have:

 $q^{n}-1=q-1+\sum_{x}\frac{q^{n}-1}{q^{n_{x}}-1}$

by last two previous properties, it results:

 $\Phi_{n}(q)\mid q^{n}-1\;,\;\Phi_{n}(q)\mid\frac{q^{n}-1}{q^{n_{x}}-1}% \Rightarrow\Phi_{n}(q)\mid q-1$

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

 $q>1\;,\;\Phi_{n}(x)\in\mathbb{Z}[x]\Rightarrow\Phi_{n}(q)\in\mathbb{Z}% \Rightarrow|\Phi_{n}(q)|\mid q-1\Rightarrow|\Phi_{n}(q)|\leqslant q-1$

If, for $n>1$,we have $|\Phi_{n}(q)|>q-1$, 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 $\mathbb{C}$

 $|q-\xi|\geqslant||q|-|\xi||=|q-1|$

as $\xi$ is a primitive root of unity, besides

 $|q-\xi|=|q-1|\Leftrightarrow\xi=1$

but

 $n>1\Rightarrow\xi\neq 1$

therefore, we have

 $|q-\xi|>|q-1|=q-1\Rightarrow|\Phi_{n}(q)|>q-1$
Title second proof of Wedderburn’s theorem SecondProofOfWedderburnsTheorem 2013-03-22 13:34:39 2013-03-22 13:34:39 Mathprof (13753) Mathprof (13753) 17 Mathprof (13753) Proof msc 12E15