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). {{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 $x$ with respect to $D$. Then we know by the Conjugacy Class Equation that: :$\displaystyle \order D = \order {\map Z D} + \sum_{j \mathop = 0}^{l - 1} \index D {\map {N_D} {x_j} }$ which by Lagrange's theorem is: :$\displaystyle \order D + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {x_j} } }$ 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 $D$. 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: {{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.   :$(1):\quad$ 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. {{refactor|The above two results need to be proved, on their own pages.}} {{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: :$\paren {q^n}^{\alpha_i} - 1 \divides \paren {q^n}^m - 1$ 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. {{qed}}