Signalizer functor

In mathematics, in the area of abstract algebra, a signalizer functor is a mapping from a potential finite subgroup to the centralizers of the nontrivial elements of an abelian group. The signalizer functor theorem provides the conditions under which the source of such a functor is in fact a subgroup.

The signalizer functor was first defined by Daniel Gorenstein.^[1] George Glauberman proved the Solvable Signalizer Functor Theorem for solvable groups^[2] and Patrick McBride proved it for general groups.^[3][4] Results concerning signalizer functors play a major role in the classification of finite simple groups.

Let A be a non-cyclic elementary abelian p-subgroup of the finite group G. An A-signalizer functor on G (or simply a signalizer functor when A and G are clear) is a mapping θ from the set of nonidentity elements of A to the set of A-invariant p′-subgroups of G satisfying the following properties:

• For every nonidentity element ${\displaystyle a\in A}$, the group ${\displaystyle \theta (a)}$ is contained in ${\displaystyle C_{G}(a).}$
• For every pair of nonidentity elements ${\displaystyle a,b\in A}$, we have ${\displaystyle \theta (a)\cap C_{G}(b)\subseteq \theta (b).}$

The second condition above is called the balance condition. If the subgroups ${\displaystyle \theta (a)}$ are all solvable, then the signalizer functor ${\displaystyle \theta }$ itself is said to be solvable.

Given ${\displaystyle \theta ,}$ certain additional, relatively mild, assumptions allow one to prove that the subgroup ${\displaystyle W=\langle \theta (a)\mid a\in A,a\neq 1\rangle }$ of ${\displaystyle G}$ generated by the subgroups ${\displaystyle \theta (a)}$ is in fact a ${\displaystyle p'}$-subgroup.

The Solvable Signalizer Functor Theorem proved by Glauberman states that this will be the case if ${\displaystyle \theta }$ is solvable and ${\displaystyle A}$ has at least three generators.^[2] The theorem also states that under these assumptions, ${\displaystyle W}$ itself will be solvable.

Several weaker versions of the theorem were proven before Glauberman's proof was published. Gorenstein proved it under the stronger assumption that ${\displaystyle A}$ had rank at least 5.^[1] David Goldschmidt proved it under the assumption that ${\displaystyle A}$ had rank at least 4 or was a 2-group of rank at least 3.^[5][6] Helmut Bender gave a simple proof for 2-groups using the ZJ theorem,^[7] and Paul Flavell gave a proof in a similar spirit for all primes.^[8] Glauberman gave the definitive result for solvable signalizer functors.^[2] Using the classification of finite simple groups, McBride showed that ${\displaystyle W}$ is a ${\displaystyle p'}$-group without the assumption that ${\displaystyle \theta }$ is solvable.^[3][4]

The terminology of completeness is often used in discussions of signalizer functors. Let ${\displaystyle \theta }$ be a signalizer functor as above, and consider the set И of all ${\displaystyle A}$-invariant ${\displaystyle p'}$-subgroups ${\displaystyle H}$ of ${\displaystyle G}$ satisfying the following condition:

• ${\displaystyle H\cap C_{G}(a)\subseteq \theta (a)}$ for all nonidentity ${\displaystyle a\in A.}$

For example, the subgroups ${\displaystyle \theta (a)}$ belong to И as a result of the balance condition of θ.

The signalizer functor ${\displaystyle \theta }$ is said to be complete if И has a unique maximal element when ordered by containment. In this case, the unique maximal element can be shown to coincide with ${\displaystyle W}$ above, and ${\displaystyle W}$ is called the completion of ${\displaystyle \theta }$. If ${\displaystyle \theta }$ is complete, and ${\displaystyle W}$ turns out to be solvable, then ${\displaystyle \theta }$ is said to be solvably complete.

Thus, the Solvable Signalizer Functor Theorem can be rephrased by saying that if ${\displaystyle A}$ has at least three generators, then every solvable ${\displaystyle A}$-signalizer functor on ${\displaystyle G}$ is solvably complete.

The easiest way to obtain a signalizer functor is to start with an ${\displaystyle A}$-invariant ${\displaystyle p'}$-subgroup ${\displaystyle M}$ of ${\displaystyle G,}$ and define ${\displaystyle \theta (a)=M\cap C_{G}(a)}$ for all nonidentity ${\displaystyle a\in A.}$ However, it is generally more practical to begin with ${\displaystyle \theta }$ and use it to construct the ${\displaystyle A}$-invariant ${\displaystyle p'}$-group.

The simplest signalizer functor used in practice is ${\displaystyle \theta (a)=O_{p'}(C_{G}(a)).}$

As defined above, ${\displaystyle \theta (a)}$ is indeed an ${\displaystyle A}$-invariant ${\displaystyle p'}$-subgroup of ${\displaystyle G}$, because ${\displaystyle A}$ is abelian. However, some additional assumptions are needed to show that this ${\displaystyle \theta }$ satisfies the balance condition. One sufficient criterion is that for each nonidentity ${\displaystyle a\in A,}$ the group ${\displaystyle C_{G}(a)}$ is solvable (or ${\displaystyle p}$-solvable or even ${\displaystyle p}$-constrained).

Verifying the balance condition for this ${\displaystyle \theta }$ under this assumption can be done using Thompson's ${\displaystyle P\times Q}$-lemma.

To obtain a better understanding of signalizer functors, it is essential to know the following general fact about finite groups:

• Let ${\displaystyle E}$ be an abelian non-cyclic group acting on the finite group ${\displaystyle X.}$ Assume that the orders of ${\displaystyle E}$ and ${\displaystyle X}$ are relatively prime.
• Then ${\displaystyle X=\langle C_{X}(E_{0})\mid E_{0}\subseteq E,{\text{ and }}E/E_{0}{\text{ cyclic }}\rangle }$

This fact can be proven using the Schur–Zassenhaus theorem to show that for each prime ${\displaystyle q}$ dividing the order of ${\displaystyle X,}$ the group ${\displaystyle X}$ has an ${\displaystyle E}$-invariant Sylow ${\displaystyle q}$-subgroup. This reduces to the case where ${\displaystyle X}$ is a ${\displaystyle q}$-group. Then an argument by induction on the order of ${\displaystyle X}$ reduces the statement further to the case where ${\displaystyle X}$ is elementary abelian with ${\displaystyle E}$ acting irreducibly. This forces the group ${\displaystyle E/C_{E}(X)}$ to be cyclic, and the result follows. ^[9][10]

This fact is used in both the proof and applications of the Solvable Signalizer Functor Theorem.

For example, one useful result is that it implies that if ${\displaystyle \theta }$ is complete, then its completion is the group ${\displaystyle W}$ defined above.

Another result that follows from the fact above is that the completion of a signalizer functor is often normal in ${\displaystyle G}$:

Let ${\displaystyle \theta }$ be a complete ${\displaystyle A}$-signalizer functor on ${\displaystyle G}$.

Let ${\displaystyle B}$ be a noncyclic subgroup of ${\displaystyle A.}$ Then the coprime action fact together with the balance condition imply that${\displaystyle W=\langle \theta (a)\mid a\in A,a\neq 1\rangle =\langle \theta (b)\mid b\in B,b\neq 1\rangle .}$

To see this, observe that because ${\displaystyle \theta (a)}$ is B-invariant, ${\displaystyle \theta (a)=\langle \theta (a)\cap C_{G}(b)\mid b\in B,b\neq 1\rangle \subseteq \langle \theta (b)\mid b\in B,b\neq 1\rangle .}$

The equality above uses the coprime action fact, and the containment uses the balance condition. Now, it is often the case that ${\displaystyle \theta }$ satisfies an "equivariance" condition, namely that for each ${\displaystyle g\in G}$ and nonidentity ${\displaystyle a\in A}$, ${\displaystyle \theta (a^{g})=\theta (a)^{g}\,}$ where the superscript denotes conjugation by ${\displaystyle g.}$ For example, the mapping ${\displaystyle a\mapsto O_{p'}(C_{G}(a))}$, the example of a signalizer functor given above, satisfies this condition.

If ${\displaystyle \theta }$ satisfies equivariance, then the normalizer of ${\displaystyle B}$ will normalize ${\displaystyle W.}$ It follows that if ${\displaystyle G}$ is generated by the normalizers of the noncyclic subgroups of ${\displaystyle A,}$ then the completion of ${\displaystyle \theta }$ (i.e., W) is normal in ${\displaystyle G.}$