Frobenius Groups

Let $G$ be a finite group acting transitively on a set $X$. We call $G$ a Frobenius group if only the identity element fixes more than one point. In other words, if $x, y$ are distinct elements, and if $g x = x$ and $g y = y$ then $g = 1$. We assume that $X$ has more than one element.

Example 3.3.1: Suppose that $F$ is a finite field, and let $X = F$. This is the affine line $\mathbb{A}^1 (F)$, which is $\mathbb{P}^1 (F) -\{\infty\}$. Let $G$ be the affine group, which is set of all affine transformations of $F$, that is, maps $g : F \longrightarrow F$ of the form $g (x) = a x + b$ for some $a, b \in F$, $a \neq 0$. If $g$ fixes $x$ and $y$ then $x = a x + b$ and $y = a y + b$. Subtracting these equations, \[ x - y = a (x - y), \qquad \text{so}\qquad a = 1, \] assuming $x \neq y$. Now $x = a x + b$ means $x = x + b$ so $b = 0$. Thus only the identity element fixes both $x$ and $y$, so this $G$ is a Frobenius group.

The above example is interesting because $G$ has a normal subgroup $K$ consisting of all translations. These are the transformation $x \longmapsto x + b$, which are the subgroup with $a = 1$.

Exercise 3.3.1: (a) Show that if we associate with $g : x \longmapsto a x + b$ the matrix \[ \left(\begin{array}{cc} a & b\\ 0 & 1 \end{array}\right) \] then we may calculate in the affine group by matrix multiplication. Discuss the relationship between the affine group and the projective group $\operatorname{PGL}_2 (F)$, and their group actions on $\mathbb{A}^1 (F)$ and $\mathbb{P}^1 (F)$.

(b) Show that $K$ is normal.

(c) Show that $K - 1$ is the union of all the isotropy subgroups of different $x \in X$.

(d) Show that if $H$ is the subgroup with $b = 0$ then $H$ is the isotropy subgroup of $0 \in X$, and $G = K \rtimes H$ (internal semidirect product).

There are other Frobenius groups, but this example is sufficient to give you some intuition for them.

There was no reference to the normal subgroup $K$ (as in the example) in the definition of a Frobenius group, but Frobenius proved that such a subgroup always exists. The proof uses character theory. This is a striking application since characters are not really involved in the statement of the theorem – just the proof.

Exercise 3.3.2: (a) Suppose that $G = H K$ is a semidirect product, so $K \vartriangleleft G$, $H K = G$ and $H \cap K = 1$. Prove that the composition

is an isomorphism.

Theorem 3.3.1: (Frobenius) Let $G$ be a Frobenius group acting on $X$. Let \[ K = 1 \cup \{g \in G| \text{$g$ has no fixed points} \}. \] Then $K$ is a normal subgroup of $G$. Moreover if $H = G_{x_0}$ is the isotropy subgroup of $x_0 \in X$, then $H \cap K = 1$, $H K = G$ and so $G = K \rtimes H$.

This is remarkable because there isn't any obvious reason why it should be true. By this definition, $K$ is just a set – why should it be closed under multiplication?

We make some remarks to motivate the proof. Let us suppose that we believe the theorem to be true but we don't know how to prove this. A common strategy is to consider the consequences of the theorem you are trying to prove, and see if you can prove those. Often once such a consequence is established, the original statement will be seen in a different light, and maybe provable.

So let us assume that $K$ is a subgroup and that $G$ is a semidirect product, as the theorem claims. Then the composition $H \longrightarrow G \longrightarrow G / K$ of the inclusion $H \longrightarrow G$ and the projection $G \longrightarrow G / K$ is an isomorphism $H \longrightarrow G / K$. This means that if $\pi : H \longrightarrow L$ is any homomorphism of $H$ to a group, then $\pi$ can be extended to $G$. Indeed, we compose $\pi$ with the inverse isomorphism $G / K \longrightarrow H$ and the composite map \[ \Pi : G \longrightarrow G / K \cong H \mathop{\longrightarrow}\limits^{\pi} L \] is a group homomorphism that extends $\pi$ to $G$.

Thus if Frobenius' theorem is true, we can extend any representation of $H$ to a representation of $G$, and it follows that we can extend any character, or generalized character of $H$ to $G$. Perhaps we can prove this consequence of Frobenius' theorem directly. Indeed, this is what we will do.

Proof. (Click to Expand/Collapse)

We will show that any generalized character $\chi$ of $H$ extends to a generalized character $\chi'$ of $G$.

Once we have this fact, we will be able to deduce Frobenius' theorem. We will first verify this statement in two special cases before tackling the general case. The two special cases are:

$\chi$ is the trivial character of $H$;
$\chi (1) = 0$;

If $\chi = 1_G$ is the trivial character, $\chi (g) = 1$ for all $g$, then just take $\chi'$ to be the trivial character of $G$. Then obviously $\chi'$ extends $\chi$, and (3.3.2) is true because both sides are $1$.

Next, suppose that $\chi (1) = 0$. Then we will take $\chi'$ to be the generalized character of $G$ induced from $\chi$. It is given by the formula
\[ \chi' (u) = \sum_{g \in G / H} \dot{\chi} (g^{- 1} u g), \] (3.3.1)

where we sum over a set of representatives for the set $G / H$ of cosets $g H$. We will show that $\chi' (h) = \chi (h)$ if $h \in H$. If $h = 1$, then $\chi' (h) = \chi (h) = 0$, so assume that $h \neq 1$. Now we know by Theorem 1.5.1 that there is a bijection $G / H \longleftrightarrow X$ in which the coset $g H$ corresponds to $g x_0$. Now let us ask how $g^{- 1} h g$ can be in $H$. This means that $g^{- 1} h g x_0 = x_0$, so $h g x_0 = g x_0$. Thus $h$ fixes both $x_0$ (since $h \in H$) and $g x_0$. Since only the identity element fixes two elements of $X$, we have $g x_0 = x_0$ and so $g \in H$. Thus only one coset representative contributes to the sum on the right-hand side of (3.3.1), namely the unique representative $g$ of the coset $H$, which we can take to be the identity. Therefore $\chi' (h) = \dot{\chi} (1^{- 1} h 1) = \chi (h)$ when $h \in H$.

In general, any $\chi$ can be written as a sum of two characters, one being a multiple of $1_G$, the other satisfying $\chi (1) = 0$, and so the general case follows from the two cases just handled. Next we show that the map $\chi \longmapsto \chi'$ satisfies
\[ \left\langle \chi_1, \chi_2 \right\rangle_H = \left\langle \chi_1', \chi_2' \right\rangle_G . \] (3.3.2)

Since this is linear in $\chi_1$ and $\chi_2$, we may handle separately the four cases:

$\chi_1 = 1_H$, $\chi_2 = 1_H$,
$\chi_1 = 1_H$, $\chi_2 (1) = 0,$
$\chi_1 (1) = 0$, $\chi_2 = 1_H$,
$\chi_1 (1) = 0$, $\chi_2 (1) = 0$.

In the case $\chi_1 = \chi_2 = 1_H$ we have $\chi_1' = \chi_2' = 1_G$, and both sides of of (3.3.2) are 1. Thus we may assume that either $\chi_1 (0) = 0$ or $\chi_2 (0) = 0$; by symmetry, we may assume that $\chi_1 (0) = 0$, and in this case $\chi_1' = \chi_1^G$. Now by Frobenius reciprocity (which is valid for generalized characters) we have \[ \left\langle \chi_1', \chi_2' \right\rangle_G = \left\langle \chi_1^G, \chi_2' \right\rangle_G = \left\langle \chi_1, \chi_2' \right\rangle_H = \left\langle \chi_1, \chi_2 \right\rangle_H, \] so (3.3.2) is proved.

Finally let us show that if $u \in K$ then
\[ \chi' (u) = \chi (1) = \chi' (1) . \] (3.3.3)

We handle the two cases $\chi = 1_G$ and $\chi (1) = 0$ separately. If $\chi = 1_G$ then both sides in this equation equal $1$ (for all $u \in G$, a fortiori all $u \in K$). On the other hand if $\chi (1) = 0$ then $\chi' (1) = 0$ and we have to show that $\chi' (u) = 0$ for $1 \neq u \in K$. We will argue that in (3.3.1) there is no way for $g^{- 1} u g$ to be in $H$. Indeed, if $g^{- 1} u g \in H$ then $g^{- 1} u g x_0 = x_0$, so $u \cdot g x_0 = g x_0$, and so $u$ fixes $g x_0$. But this can't happen. Indeed, since $1 \neq u \in K$, $u$ has no fixed points by the definition of $K$. So $\dot{\chi} (g^{- 1} u g)$ is always zero. Thus $\chi' (u) = 0$ for all $u \in K$.

Now let $(\pi, V)$ be an irreducible representation of $H$. We will prove that $\pi$ can be extended to a irreducible representation of $G$. Let $\chi'$ be the generalized character of $G$ extending the character $\chi$ of $\pi$. Then $\left\langle \chi', \chi' \right\rangle_G = \left\langle \chi, \chi \right\rangle_H = 1$ and $\chi' (1) = \chi (1) > 0$, so $\chi'$ is the character of an irreducible representation of $G$ by Lemma 2.7.2. Let $(\pi', V')$ be the irreducible representation of $G$ with character $\chi'$. Then the restriction of $\pi'$ to $H$ is isomorphic to $\pi$, which means that there is an isomorphism $f : V \longrightarrow V'$ such that \[ f (\pi (g) v) = \pi' (g) f (v) . \] Identifying $V$ with $V'$ by means of this isomorphism makes $\pi$ agree with the restriction of $V'$ to $H$, so $\pi'$ extends $\pi$.

Now, if $(\pi, V)$ is any representation of $H$ we can extend $\pi$ to a representation of $G$. Indeed, we have just checked this if $\pi$ is irreducible, then it can be extended; and in general we can decompose it into irreducible representations, and extend each of these individually.

Now we extend the regular representation $\pi_{\operatorname{reg}}$ of $H$ to a representation $\pi'_{\operatorname{reg}}$ of $G$. Let $\rho$ be the characer of $\pi_{\operatorname{reg}}$, and $\rho'$ the character of the extension. We will show that the kernel of $\pi'_{\operatorname{reg}}$ is precisely $K$ – which will show that $K$ is a normal subgroup of $G$. By Exercise 3.2.1, what we must show is that $\rho' (u) = \rho (1)$ if and only if $u \in K$. If $u \in K$, then $\rho' (u) = 1$ by (3.3.3). On the other hand, if $u \notin K$ then $u$ is conjugate to a nonidentity element $h$ of $H$, and $\rho' (u) = \rho' (h) = \rho (h) = 0$ by Theorem 2.6.2. This completes the proof.

If $G$ is a Frobenius group, with notation as in the last Theorem, we call $K$ the Frobenius kernel and $H$ the Frobenius complement.

Exercise 3.3.3: Let $G$ be a Frobenius group with kernel $K$ and complement $H$. Suppose that $h \in H$ and $1 \neq k \in K$. Prove that $h k h^{- 1} \neq k$. (Hint: if $h$ and $k$ commute, prove that $h$ fixes $k x_0$, where $x_0$ is as in Theorem 3.3.1.)

Exercise 3.3.4: Use the last exercise to prove that $|H|$ divides $|K| - 1$. Hint: let $H$ act on $K$ by conjugation. What does the last exercise say about the orbits?)

Exercise 3.3.5: Let $H$ and $K$ be finite groups, and let $\theta : H \longrightarrow \operatorname{Aut} (K)$ be a homomorphism. Assume that whenever $1 \neq h \in H$ and $1 \neq k \in K$ we have $\theta (h) k \neq k$. Show that the semidirect product $K \ltimes_{\theta} H$ can be realized as a Frobenius group with kernel $K$ and complement $H$. (Hint: Identify $H$ and $K$ with their images in $G$, so the hypothesis means that $h k h^{- 1} \neq k$. Let $G$ act on the coset space $G / H$. Observe that every coset $k H$ has a unique representative $k \in K$. To prove that this is a Frobenius group, suppose that $1 \neq g \in G$ has two fixed points $k H$ and $k' H$. Show that $h = k^{- 1} g k \in H$. Then show that $h k'' H = k'' H$ where $k'' = k^{- 1} k'$. Now figure out what the hypothesis tells you if $k'' \neq 1$.)

Exercise 3.3.6: We've only given limited examples of Frobenius groups, so you might be wondering if the kernel is always abelian. Here's a construction of a Frobenius group with nonabelian kernel. By the last exercise, you need only construct a nonabelian group $K$ and a group $H$ with a homomorphism $\theta : H \longrightarrow \operatorname{Aut} (K)$ such that if $h \neq 1$ and $k \neq 1$ then $\theta (h) k \neq k$. Take $K$ to be the group of order $7^3$ consisting of $3 \times 3$ matrices \[ \left\{ \left(\begin{array}{ccc} 1 & x & y\\ & 1 & z\\ & & 1 \end{array}\right) | x, y, z \in \mathbb{Z}/ 7\mathbb{Z} \right\} . \] Let $\sigma : K \longrightarrow K$ be the map \[ \sigma \left(\begin{array}{ccc} 1 & x & y\\ & 1 & z\\ & & 1 \end{array}\right) = \left(\begin{array}{ccc} 1 & 2 x & 4 y\\ & 1 & 2 z\\ & & 1 \end{array}\right) . \] Check that $\sigma$ is an automorphism of order 3, and explain why this solves the problem.