Theorem: Wedderburn's 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 $\conjclass {x_j}$ (the representative does not matter).
{{finish|Invoke Normalizer of Conjugate is Conjugate of Normalizer to formalise the independence of representative choice}}
Let there be $l$ (distinct) non-singleton conjugacy classes.
Let $\map {N_D} x$ denote the normalizer of $x$ {{WRT}} $D$.
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 $D$.
{{explain|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)}}
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 conjugacy 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:
:$\order {\map {N_D} u} = \paren {p^n}^r$
for $r < m$.
Suppose there are $l$ such $u$.
Then:
{{begin-eqn}}
{{eqn | l = \order {\map U D}
      | r = \order {\map Z {\map U D} } - 1 + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {u_j} } }
      | c = 
}}
{{eqn | r = p^n - 1 + \sum_{\alpha_i} \frac {\paren {p^n}^m - 1} {\paren {p^n}^{\alpha_i} - 1}
      | c = 
}}
{{end-eqn}}
{{stub|Clean the following up.  This bit is due to {{AuthorRef|Ernst Witt}}.}}
We need two results to finish.
:$(1): \quad p^k - 1 \divides p^j - 1 \implies k \divides j$
:$(2): \quad j \divides k \implies \Phi_n \divides \dfrac {x^j - 1} {x^k - 1}$
where:
:$\divides$ denotes divisibility of integer or of polynomial
:$\Phi_n$ denotes the $n$th cyclotomic polynomial.
{{ExtractTheorem|The above two results need to be proved, on their own pages.<br/>
$(1)$: Integer Less One divides Power Less One/Corollary/Converse}}
{{AimForCont}} $m > 1$.
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:
:$\paren {q^n}^{\alpha_i} - 1 \divides \paren {q^n}^m - 1$
From $(1)$:
:$\alpha_i \divides m$
From $(2)$:
:$\Phi_m \divides \dfrac {x^m - 1} {x^{\alpha_i} - 1}$
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.
{{qed}}