# Wedderburn's Theorem

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## Theorem

Every finite division ring $D$ is a field.

## Proof

Let $D$ be a finite division ring.

If $D$ is shown commutative then, by definition, $D$ is a field.

Let $\map Z D$ be the center of $D$, that is:

- $\map Z D := \set {z \in D: \forall d \in D: z d = d z}$

From Center of Division Ring is Subfield it follows that $\map Z D$ is a Galois field.

Thus from Characteristic of Galois Field is Prime the characteristic of $\map Z D$ is a prime number $p$.

Let $\Z / \ideal p$ denote the quotient ring over the principal ideal $\ideal p$ of $\Z$.

From Field of Prime Characteristic has Unique Prime Subfield, the prime subfield of $\map Z D$ is isomorphic to $\Z / \ideal p$.

From Division Ring is Vector Space over Prime Subfield, $\map Z D$ is thus a vector space over $\Z / \ideal p$.

From Vector Space over Division Subring is Vector Space, $D$ is a vector space over $\map Z D$.

Since $\map Z D$ and $D$ are finite, both vector spaces are of finite dimension.

Let $n$ and $m$ be the dimension of the two vector spaces respectively.

It now follows from Cardinality of Finite Vector Space that $\map Z D$ has $p^n$ elements and $D$ has $\paren {p^n}^m$ elements.

Now the idea behind the rest of the proof is as follows.

We want to show $D$ is commutative.

By definition, $\map Z D$ is commutative.

Hence it is to be shown that $D = \map Z D$.

It is shown that:

- $\order D = \order {\map Z D}$

Hence $D = \map Z D$, and the proof is complete.

$\map Z D$ and $D$ are considered as modules.

We have that if $m = 1$ then:

- $\order D = \order {\map Z D}$

and the result then follows.

Thus it remains to show that $m = 1$.

In a finite group, let $x_j$ be a representative of the conjugacy class $\tuple {x_j}$ (the representative does not matter).

This needs considerable tedious hard slog to complete it.In particular: Invoke Normalizer of Conjugate is Conjugate of Normalizer to formalise the independence of representative choiceTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Let there be $l$ (distinct) non-singleton conjugacy classes.

Let $\map {N_D} x$ denote the normalizer of $x$ with respect to $D$.

Then we know by the Conjugacy Class Equation that:

- $\ds \order D = \order {\map Z D} + \sum_{j \mathop = 0}^{l - 1} \index D {\map {N_D} {x_j} }$

which by Lagrange's theorem is:

- $\ds \order D + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {x_j} } }$

Consider the group of units $\map U D$ in $D$.

Consider what the above equation tells if we start with $\map U D$ instead of $D$.

This article, or a section of it, needs explaining.In particular: We cannot take $D$ in the first place, since $D$ is not a group under multiplication. Doesn't it make sense to start with $\map U D$ directly? --Wandynsky (talk) 16:51, 30 July 2021 (UTC)You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

If we centralize a multiplicative unit that is in the center, from Conjugacy Class of Element of Center is Singleton we get a singleton conjugacy class.

Bear in mind that the above sum only considers non-singleton classes.

Thus choose some element $u$ not in the center, so $\map {N_D} u$ is not $D$.

However, $\map Z D \subset \map {N_D} u$ because any element in the center commutes with everything in $D$ including $u$.

Then:

- $\order {\map {N_D} u} = \paren {p^n}^m$

for $r < m$.

Suppose there are $l$ such $u$.

Then:

\(\ds \order {\map U D}\) | \(=\) | \(\ds \order {\map Z {\map U D} } - 1 + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {u_j} } }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds p^n - 1 + \sum_{\alpha_i} \frac {\paren {p^n}^m - 1} {\paren {p^n}^{\alpha_i} - 1}\) |

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We need two results to finish.

- $(1):\quad$ If $p^k - 1 \divides p^j - 1$, then $k \divides j$

where $\divides$ denotes divisibility.

- $(2)\quad$ If $j \divides k$ then $\Phi_n \divides \dfrac{x^j - 1} {x^k - 1}$

where $\Phi_n$ denotes the $n$th cyclotomic polynomial.

This page or section has statements made on it that ought to be extracted and proved in a Theorem page.In particular: The above two results need to be proved, on their own pages.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by creating any appropriate Theorem pages that may be needed.To discuss this page in more detail, feel free to use the talk page. |

Aiming for a contradiction, suppose $m > 1$.

Let $\gamma_i$ be an $m$th primitive root of unity.

Then the above used conjugacy class theorem tells us how to compute size of $\map U D$ using non-central elements $u_j$.

However, in doing so, we have that:

- $\paren {q^n}^{\alpha_i} - 1 \divides \paren {q^n}^m - 1$

Thus by the first result:

- $\alpha_i \divides m$

Thus:

- $\Phi_m \divides \dfrac {x^m - 1} {x^{\alpha_i} - 1}$

However:

- $\size {p^n - \gamma_i} > p^n - 1$

Thus the division is impossible.

This contradicts our assumption that $m > 1$.

Hence $m = 1$ and the result follows, as determined above.

$\blacksquare$

## Also known as

**Wedderburn's Theorem** is also known as **Wedderburn's Little Theorem**.

## Source of Name

This entry was named for Joseph Henry Maclagan Wedderburn.

## Historical Note

Wedderburn first published the result now known as **Wedderburn's Theorem** in $1905$.

However, his proof had a gap in it.

The first complete proof was supplied by Leonard Eugene Dickson.