A finite group is said to be a Frobenius group if there is a non-trivial subgroup of (known as the Frobenius complement of ) such that the conjugates of are “disjoint as possible” in the sense that whenever . This gives a decomposition
where the Frobenius kernel of is defined as the identity element together with all the non-identity elements that are not conjugate to any element of . Taking cardinalities, we conclude that
A remarkable theorem of Frobenius gives an unexpected amount of structure on and hence on :
Theorem 1 (Frobenius’ theorem) Let be a Frobenius group with Frobenius complement and Frobenius kernel . Then is a normal subgroup of , and hence (by (2) and the disjointness of and outside the identity) is the semidirect product of and .
I discussed Frobenius’ theorem and its proof in this recent blog post. This proof uses the theory of characters on a finite group , in particular relying on the fact that a character on a subgroup can induce a character on , which can then be decomposed into irreducible characters with natural number coefficients. Remarkably, even though a century has passed since Frobenius’ original argument, there is no proof known of this theorem which avoids character theory entirely; there are elementary proofs known when the complement has even order or when is solvable (we review both of these cases below the fold), which by the Feit-Thompson theorem does cover all the cases, but the proof of the Feit-Thompson theorem involves plenty of character theory (and also relies on Theorem 1). (The answers to this MathOverflow question give a good overview of the current state of affairs.)
I have been playing around recently with the problem of finding a character-free proof of Frobenius’ theorem. I didn’t succeed in obtaining a completely elementary proof, but I did find an argument which replaces character theory (which can be viewed as coming from the representation theory of the non-commutative group algebra ) with the Fourier analysis of class functions (i.e. the representation theory of the centre of the group algebra), thus replacing non-commutative representation theory by commutative representation theory. This is not a particularly radical depature from the existing proofs of Frobenius’ theorem, but it did seem to be a new proof which was technically “character-free” (even if it was not all that far from character-based in spirit), so I thought I would record it here.
The main ideas are as follows. The space of class functions can be viewed as a commutative algebra with respect to the convolution operation ; as the regular representation is unitary and faithful, this algebra contains no nilpotent elements. As such, (Gelfand-style) Fourier analysis suggests that one can analyse this algebra through the idempotents: class functions such that . In terms of characters, idempotents are nothing more than sums of the form for various collections of characters, but we can perform a fair amount of analysis on idempotents directly without recourse to characters. In particular, it turns out that idempotents enjoy some important integrality properties that can be established without invoking characters: for instance, by taking traces one can check that is a natural number, and more generally we will show that is a natural number whenever is a subgroup of (see Corollary 4 below). For instance, the quantity
is a natural number which we will call the rank of (as it is also the linear rank of the transformation on ).
is an integer. On the other hand, one can also show by elementary means that this quantity lies between and . These two facts are not strong enough on their own to impose much further structure on , unless one restricts attention to minimal idempotents . In this case spectral theory (or Gelfand theory, or the fundamental theorem of algebra) tells us that has rank one, and then the integrality gap comes into play and forces the quantity (3) to always be either zero or one. This can be used to imply that the convolution action of every minimal idempotent either preserves or annihilates it, which makes itself an idempotent, which makes normal.
— 1. Idempotent theory —
Let be a finite group. Then we can form the Hilbert space of complex functions with inner product
where we use the usual averaging notation . This Hilbert space is also a complex *-algebra using the adjoint map
and the convolution product
For future reference, we record the adjoint relations
for any .
As an algebra, is isomorphic to the group algebra of , if one identifies each group element with the “Dirac delta function” at , defined by
for any and .
The convolution algebra will in general be non-commutative. However, it contains a commutative (in fact central) *-subalgebra , namely the space of class functions, that is to say the functions which are conjugation invariant in the sense that
for all . It is easy to see that is a central *-subalgebra of , thus if are class functions then so is and , and one has for any . Indeed it is not hard to show that is the centre of , although we will not need this fact here.
Let us call a class function an idempotent if . Thus for instance , , and are idempotents, and the convolution of any two idempotents is again an idempotent. Note that if is idempotent, then the linear transformation on is also idempotent and also self-adjoint (since and are class functions and thus commute with each other), and thus is an orthogonal projection onto the space . The projection similarly projects to the subspace . We define the rank of the idempotent to be the rank of on , or equivalently the dimension of the space . Thus for instance has rank , has rank equal to the number of conjugacy classes of , and has rank one.
We place an ordering on idempotents by declaring if , thus for instance , and for any idempotents . The relation is equivalent to the range of the projection being a subspace of the range of the projection . In particular this shows that is a partial order on idempotents, and that whenever . If , we see that is also an idempotent with . From this we see that every non-zero idempotent is either minimal (in the sense that there are no other non-zero idempotents such that , or can be split up as the sum of two strictly smaller idempotents. From this we see that we can decompose into the sum of minimal idempotents , whose ranges then form an orthogonal decomposition of (and similarly the form an orthogonal decomposition of ); and by convolving this fact with any other idempotent we see that all other idempotents take the form for some subset of . Conversely, one can show that is an idempotent for every . Thus, the structure of idempotents is completely determined by the set of minimal idempotents.
The theory thus far relied only on elementary algebra; in particular, one could replace the complex numbers here by any other field of characteristic not a factor of and still get essentially an identical theory (replacing the complex conjugate by some other automorphism, such as the trivial automorphism). But the next fact is less elementary (being analogous to Plancherel’s theorem for finite groups) and relies crucially on the algebraically closed nature of (or closely related facts, such as Liouville’s theorem).
Proof: Let be an idempotent of rank greater than one, then contains a function which is not a scalar multiple of . The linear transformation is a normal transformation on . Since is not a multiple of , is not a multiple of the identity, thus by the spectral theorem there is an eigenspace which is a proper non-trivial subspace of . This space is a convolution ideal of ; if we let be the orthogonal projection from to , then commutes with convolutions and in particular for all . Applying this to we conclude that is an idempotent, with , thus is not minimal. The claim follows (note that minimal idempotents by definition cannot be of rank zero).
Now we record some integrality properties of idempotents. We have the following trivial but crucial fact:
Lemma 3 Let be a finite-dimensional vector space, and let be an idempotent linear transformation. Then the trace of is equal to the dimension of the range of . In particular, the trace is a natural number.
Proof: This follows by computing the trace in a basis of together with a basis of some complementary subspace of .
is a natural number.
Proof: We introduce the left and right translation operators
on for , together with the projection operator
introduced previously. Observe that the three operators commute with each other. As such, the operator
is idempotent. A short computation shows that the trace of is equal to
and the claim now follows from Lemma 3.
— 2. Proof of Frobenius’ theorem —
We now begin the proof of Frobenius’ theorem. We begin with an elementary consequence of Corollary 4:
is an integer.
Proof: When the quantity (4) vanishes, so we may assume (by the decomposition ) that the idempotent has mean zero.
Applying Corollary 4 to the subgroups , , and of , we conclude that the expressions
are natural numbers (and hence integers). We now consider various linear combinations of these quantities. We write (6) as
Subtracting this from (5), we conclude that
is an integer; comparing this with (4), we see that it suffices to show that
is an integer.
If we multiply (8) by and isolate the terms where or is equal to , we obtain
Subtracting two copies of (7), we conclude that
is an integer. Using the conjugation invariance of in and , together with (1), we can rewrite this expression as
On the other hand, as has mean zero, we have
for any , and similarly
for any . Thus we can rewrite the previous expression as
On the other hand, we have an alternate representation for the quantity (4):
Lemma 6 Let be a finite group with Frobenius complement and Frobenius kernel . Then for any idempotent in , the quantity (4) is equal to , where is the orthogonal projection to the space of class functions of mean zero supported on . In particular, since is a subspace of , the quantity (4) lies between and .
Proof: The quantity can be expanded as
which can be expanded further as
This simplifies to
and a routine computation (using the conjugation invariance of ) shows that this is the same quantity as (4).
Let be a finite group with Frobenius complement and Frobenius kernel , and let be a minimal idempotent of . By Lemma 3, Lemma 5, and Lemma 6, we see that the quantity (4) is an integer between and , and is thus exactly or exactly . From Lemma 6, we thus conclude that the space either contains the range of , or is orthogonal to this range. Summing over minimal idempotents, we conclude that is the direct sum of the ranges of some minimal idempotents ; the same is thus true of the orthogonal complement of in , that is to say the space of class functions that are constant on . In particular, this space is closed under convolution, which implies that is constant on . However, this convolution attains a maximum value of at the set , and hence , thus is a subgroup. As was already known to be closed under conjugation, Theorem 1 follows.
Remark 1 As mentioned previously, one can use character theory to describe the minimal idempotents as the functions of the form , where are the irreducible characters of . The integrality property in Corollary 4 can then be obtained by taking the representation of associated to and decomposing it into irreducible representations of ; we omit the details. Thus one can view the above argument as being a disguised version of the usual character-theoretic proof. However, the point is that one can prove Corollary 4 without knowing the relationship between minimal idempotents, irreducible characters, and irreducible representations. (Indeed, Corollary 4 seems to be weaker than the fact from character theory that the restriction or induction of a character is again a character (and in particular decomposes into a natural number combination of irreducible characters); see this MathOverflow post for some further discussion.)
— 3. The solvable case —
In the case when the Frobenius complement is solvable, an elementary proof of Theorem 1 was discovered by Grun and by Shaw, based on the transfer homomorphism. To describe this argument, we first consider the model case when is abelian. We then enumerate the right cosets of as for some . This enumeration is not unique; in addition to permutations of the , we have the freedom to arbitrarily multiply each on the right by an element of to obtain without affecting the coset: .
Regardless of this freedom, we can define the transfer homomorphism for by noting that for each coset , one has for some unique depending on and , thus for some depending on , and . We then set
One observes that this map is in fact independent of the choice of coset enumeration ; permutations of are irrelevant as is assumed to be abelian, and shifting one of the by an element of causes one of the to be multiplied by , byt another of the to be divided by , leaving unchanged. Once one knows that is independent of the choice of coset enumeration, it is not difficult to verify the homomorphism property , for instance by using one enumeration to compute and , and another enumeration to compute .
We now make the crucial observation that if is an abelian Frobenius complement, then the transfer map is the identity on , thus for all . To see this, let be the order of . We claim that the action of on the quotient space consists of a single fixed point at , together with orbits of size exactly . Indeed, is clearly fixed by since , and if we have for some coset and some , then lies in and so is a multiple of , whence the claim. If we then use a coset enumeration that consists of together with -tuples of the form for each orbit of , we see that as claimed.
As is a surjective homomorphism from to , the kernel is a normal subgroup of order ; also, as is the identity on , this kernel is disjoint from , and hence also from every conjugate of . From (1) we conclude that the kernel is , and hence is normal as required.
Now we adapt the above argument to the case when is solvable instead of abelian. Here, one can still define a transfer map , but it takes values in the abelianisation of , defined by
One can verify as before that is independent of the choice of enumeration, and is a homomorphism. The previous argument also shows that the restriction of to is the quotient map from to . Thus, the kernel is a normal subgroup of whose order is and which intersects in precisely . From this we see that is the union of and the conjugates of , and so is also a Frobenius group with Frobenius complement and Frobenius kernel . The claim then follows by an induction of the derived length of .
Remark 2 This argument in fact shows that to prove Frobenius’s theorem, it suffices to do so in the case that is a perfect group.
— 4. The even order case —
When is even, Cauchy’s theorem shows that contains at least one non-trivial involution – an element of order two, since otherwise would split up into the identity and pairs , contradicting the even nature of . One can then analyse the action of these involutions combinatorially; this was first done by Bender (and we use an argument from this text of Passman). Indeed, if is an involution in , then as discussed in the previous section, the action of on fixes and breaks the rest of into orbits of size two; in particular, is odd, and there are pairs of cosets in that are swapped by . More generally, any involution in a conjugate of will fix and then swap pairs among the other cosets of .
Suppose that contains at least two involutions; then by (1) contains at least involutions, each of which swaps pairs in . But there are only such pairs; thus by the pigeonhole principle there are two cosets, say and , which are swapped by two different involutions in , thus and . But this implies that fixes both and , and thus lies in , and is thus trivial by (1), contradicting the disjoint nature of and . Thus contains exactly one involution. The same argument then shows that cannot contain any involutions. Thus the set of involutions has order exactly , consisting of the involution of and all of its conjugates.
If are involutions, then by the above discussion and lie in different conjugates of , and so the actions and fix different cosets of . Also, we have seen that they swap different pairs of cosets. As such, the combined action cannot fix any elements of , and thus by (1) lies in . We conclude that ; since and have the same order, this implies that for all , and hence . This makes a group (and thus a normal group, since it is conjugation invariant), as desired.
Remark 3 In this setting we can even establish the additional fact that is abelian. Indeed, if is an involution, then , so for every , is an involution, which implies that . Thus the inversion map is a homomorphism on , which forces to be abelian.