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 ${\mathrm{\Phi }}_d(q)$ the d-th cyclotomic polynomial.
It results

• ${\mathrm{\Phi }}_n(q)={\prod }_{\xi \in P_n}(q-\xi )$

• $p(q)=q^n-1={\prod }_{\xi \in G_n}(q-\xi )={\prod }_{d\mid n}{\mathrm{\Phi }}_d(q)$

• ${\mathrm{\Phi }}_n(q)\in \mathbb{Z}[q]$, it has multiplicative identity and ${\mathrm{\Phi }}_n(q)\mid q^n-1$

• ${\mathrm{\Phi }}_n(q)\mid \frac{q^n-1}{q^d-1}$with

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:

$${\mathrm{\Phi }}_n(q)\mid q^n-1,{\mathrm{\Phi }}_n(q)\mid \frac{q^n-1}{q^{n_x}-1}\Rightarrow {\mathrm{\Phi }}_n(q)\mid q-1$$

because ${\mathrm{\Phi }}_n(q)$ 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

$$q>1,{\mathrm{\Phi }}_n(x)\in \mathbb{Z}[x]\Rightarrow {\mathrm{\Phi }}_n(q)\in \mathbb{Z}\Rightarrow |{\mathrm{\Phi }}_n(q)|\mid q-1\Rightarrow |{\mathrm{\Phi }}_n(q)|⩽q-1$$

If, for $n>1$,we have $|{\mathrm{\Phi }}_n(q)|>q-1$, then $n=1$ and the theorem is proved.
We know that

$$|{\mathrm{\Phi }}_n(q)|=\prod _{\xi \in P_n}|q-\xi |,withq-\xi \in ℂ$$

by the triangle inequality in $ℂ$

$$|q-\xi |⩾||q|-|\xi ||=|q-1|$$

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

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

but

$$n>1\Rightarrow \xi \ne 1$$

therefore, we have

$$|q-\xi |>|q-1|=q-1\Rightarrow |{\mathrm{\Phi }}_n(q)|>q-1$$

Title	second proof of Wedderburn’s theorem
Canonical name	SecondProofOfWedderburnsTheorem
Date of creation	2013-03-22 13:34:39
Last modified on	2013-03-22 13:34:39
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	17
Author	Mathprof (13753)
Entry type	Proof
Classification	msc 12E15