Wedderburn’s theorem

[This post is part of a series on the long-standing conjecture that all finite projective planes have prime power order. See here for a list of the other posts. ]

Recall that a “division ring” is a field except that the multiplication is not necessarily commutative. The theorem is that any finite division ring is a field.

The theorem was first proved in 1905. When I was young, MacLagan Wedderburn got the credit because his proof was published first. But there have since been claims (which I have not checked) that his proof was wrong, so that the credit should go to Leonard Dickson, a colleague at Princeton, who published another proof soon afterwards.

So let $$D$$ be a finite division ring. For any element $$g\in D$$ let $$C(g)$$, the centraliser be $$\{d\in D:dg=gd\}$$. The centre $$Z$$ is the set of elements which commute with all elements of $$D$$. Evidently $$\{0,1\}\subseteq C(G)\subseteq D$$, and $$C(g)$$ is a division ring, whilst $$Z$$ is a field.

Let $$k=|Z|$$, so $$k$$ is an integer $$\ge 2$$. Let $$S=\{a_1,\dots,a_n\}$$ be (one of ) the smallest sets of elements such that every element of $$D$$ can be written as a linear combination $$\sum z_ia_i$$ with all $$z_i\in Z$$. Such sets clearly exist, for example, we could take $$S=D$$. All of the $$k^n$$ linear combinations must be distinct, for otherwise we would have $$\sum z_ia_i=\sum z’_ia_i$$ and $$z_j\ne z’_j$$ for some $$j$$ but then, subtracting $$(z_j-z’_j)a_j$$ is a linear combination of other $$a_i$$, and multiplying through by $$(z_j-z’_j)^{-1}$$, we have that $$a_j$$ is a linear combination of the other $$a_i$$ so we could get a set smaller than $$S$$ by deleting $$a_j$$. Contradiction. Hence $$D$$ has $$k^n$$ elements. Similarly, $$C(g)$$ has $$k^m$$ elements for some $$m\le n$$. If $$D$$ is not a field, then for some $$g$$ we have $$C(g)\ne D$$ and hence $$m\lt n$$.

Recall that for any group $$G$$ we can define a conjugacy relation: $$g$$~$$h$$ if and only if $$g=a^{-1}ha$$ for some $$a$$ in the group. It is easy to check that this is an equivalence relation, so that it divides the group into disjoint “conjugacy classes”. The elements in the centre $$Z$$ are precisely those elements that belong to a conjugacy class with just one element. Hence we have: $$|G|=|Z|+\sum |K_i|$$, where the sum is taken over all conjugacy classes with at least two elements. That is known as the class equation. Moreover, it is easy to check that $$|K_i|$$ is $$|G|/|C(g)|$$ where $$g\in K_i$$.

Thus in this case we have $$|k^n-1|=(k-1)+\sum{\frac {k^n-1}{k^m_i-1}}$$ where the sum is taken over the conjugacy classes with at least two elements and hence $$m_i\lt n$$. If we write $$n=qm+r$$ with $$0\le r\lt m$$ then $$k^n-1=(k^m-1)(k^{m(q-1)}+\dots+1)k^r+k^r-1$$, so if $$k^m-1$$ divides $$k^n-1$$ then it also divides $$k^r-1$$ which is impossible unless $$r=0$$. So each $$m_i$$ must divide $$n$$.

Now we know from elementary results on cyclotomic polynomials $$\Phi_n(x)\in\mathbb{Z}[x]$$ that $$\Phi_n(x)$$ is a factor of $$x^n-1$$ and of $$(x^n-1)/(x^m-1)$$ for $$m\lt n$$ and $$m|n$$ (see Exercise 7 for the latter). We may set $$x$$ to be the integer $$k$$ and we now have as relations between integers that $$\Phi_n(k)$$ is a factor of $$k^n-1$$ and of $$(k^n-1)/(k^{m_i}-1)$$. Hence the class equation implies that it is also a factor of $$k-1$$.

Suppose that $$z=a+ib$$ is any (complex) root of unity of order $$n$$. Then $$|z|=1$$ so $$|a|\lt 1$$. Hence $$|k-z|^2=k^2-2ak+1\gt k^2-2k+1=(k-1)^2\gt 1$$. Hence $$|\Phi_n(k)|=\prod_z|k-z|\gt k-1$$. Contradiction. $$\Box$$