Challenging Intuition in Modular Representation Theory: The Divisibility of Irreducible Representation Dimensions — Epoche C1
The group $\mathrm{SL}_2(\mathbb{F}_7)$ — the invertible $2\times 2$ matrices of determinant $1$ with entries in the field of seven elements — has order $336$, and over a field of characteristic $7$ it possesses an irreducible representation of dimension $5$. Since $5$ does not divide $336$, one of the most useful facts of ordinary representation theory is simply false here. That fact is Frobenius's divisibility theorem: over the complex numbers, the dimension of every irreducible representation of a finite group $G$ divides $|G|$. This essay establishes what does and does not survive when the field has characteristic $p$ dividing $|G|$, and in the course of doing so it must withdraw a claim made in the earlier version of the essay, which asserted a theorem that is not true. The vocabulary first, since everything below depends on it. A representation of a finite group $G$ over a field $K$ is a vector space $V$ over $K$ together with an action of $G$ by invertible linear maps; equivalently, $V$ is a module over the group algebra $K[G]$, the $|G|$-dimensional algebra whose basis is the elements of $G$ and whose multiplication extends that of $G$. A representation is irreducible , or the module simple , when its only $G$-stable subspaces are $0$ and itself. When $\operatorname{char} K = p$ and $p$ divides $|G|$ the subject is called modular representation theory. Why characteristic zero is well behaved The whole of the classical theory rests on one theorem, and seeing where its proof uses a division makes the modular failure predictable rather than surprising. Maschke's theorem states that if $\operatorname{char} K$ does not divide $|G|$, then every $K[G]$-module is a direct sum of simple ones — the algebra is semisimple . The proof takes any linear projection $\pi$ of $V$ onto a submodule $W$ and averages it over the group, $$\tilde{\pi} = \frac{1}{|G|}\sum_{g \in G} g\,\pi\, g^{-1},$$ which is a $G$-equivariant projection onto $W$, so that $V = W \oplus \ker\tilde{\pi}$ splits as a module. Every step is elementary except one: the factor $1/|G|$ requires $|G|$ to be invertible in $K$. Two consequences follow for $K$ algebraically closed of characteristic zero. By Wedderburn's structure theorem a semisimple algebra is a product of matrix algebras, one for each simple module, so comparing dimensions gives $$\sum_{i=1}^{k}(\dim V_i)^2 = |G|,$$ with $k$ the number of conjugacy classes. And Frobenius proved separately that each $\dim V_i$ divides $|G|$; a proof using algebraic integers is given as Theorem 3.11 of Isaacs (1976). The earlier version of this essay presented the sum-of-squares identity as capturing the divisibility statement. It does not: the two are independent results, and the identity is consistent with dimensions that divide nothing in particular. It is also worth recording that the earlier text spoke of ordinary representations having dimensions "not divisible by the characteristic of the field", which is not a statement at all when the characteristic is zero. What breaks when $p$ divides $|G|$ The failure is not a subtlety requiring machinery; it can be exhibited in one element. Let $\operatorname{char} K = p$ divide $|G|$ and set $N = \sum_{g \in G} g$ in $K[G]$. Then $hN = N = Nh$ for every $h \in G$, so $K N$ is a two-sided ideal of dimension $1$, and $$N^2 = \sum_{h \in G} hN = |G|\,N = 0$$ because $|G|$ is zero in $K$. So $K[G]$ contains a nonzero nilpotent ideal and cannot be semisimple. What obstructs the theory is therefore the Jacobson radical $J(K[G])$, the largest nilpotent ideal, which vanishes exactly when the algebra is semisimple. Modules no longer decompose into simple pieces; they merely have composition series, and the dimension count above must be replaced by $$\dim_K \big(K[G]/J(K[G])\big) = \sum_{i}(\dim V_i)^2,$$ valid whenever $K$ is a splitting field, since $K[G]/J$ is semisimple with the same simple modules. Take $G = S_3$, the symmetric group on three letters, and $K = \mathbb{F}_2$. Here $S_3 \cong \mathrm{GL}_2(\mathbb{F}_2)$, and the natural action on $\mathbb{F}_2^2$ is irreducible of dimension $2$: the only proper nonzero subspaces are the three lines, and the group permutes them transitively, so none is stable. Together with the trivial module these are all the simple modules, for reasons given below, and $$\dim_K\big(K[S_3]/J\big) = 1^2 + 2^2 = 5 \ne 6 = |S_3|,$$ so $\dim J = 1$. In fact $J = KN$: the ideal $KN$ is nilpotent, hence contained in $J$, and both are one-dimensional. The entire gap between the semisimple picture and the truth is, in this example, the single element $N$. The claim that must be withdrawn The earlier version stated a theorem in the following form: if $G$ is finite and $K$ is algebraically closed of characteristic $p$ greater than zero, then there exists an irreducible $K[G]$-module whose dimension is divisible by $p$. This is false, and two counterexamples show it is not false only marginally. First, let $G$ be a $p$-group and $K$ a finite field of characteristic $p$. Then the only simple $K[G]$-module is the trivial one, of dimension $1$. The argument is a counting argument. Let $V$ be a nonzero finite-dimensional $K[G]$-module and regard $V$ merely as a finite set on which $G$ acts; its cardinality $|K|^{\dim V}$ is a power of $p$. Every orbit has size dividing $|G|$ and so is a power of $p$, and the orbits of size greater than one contribute multiples of $p$; hence the number of fixed points is congruent to $|V|$, and therefore to $0$, modulo $p$. The vector $0$ is fixed, so there are at least $p$ fixed points and thus a nonzero $G$-fixed vector. A simple module therefore contains a trivial submodule and equals it. The same conclusion holds over any field of characteristic $p$. The earlier text noticed this case and set it aside on the ground that "for a general group $G$ where $p$ divides $|G|$ the situation changes", but $p$ divides the order of a $p$-group, so the case is not excluded by the hypothesis as stated