Centralizer product theorem

Statement

For faithful group actions

Suppose $p,q$ are distinct primes. Let $P$ be a finite $p$ -group, and $Q\leq \operatorname {Aut} (P)$ be an abelian $q$ -group that is not cyclic. Suppose $x_{1},x_{2},\dots ,x_{n}$ are the non-identity elements of $Q$ , enumerated in any arbitrary order. Then, we have:

$P=C_{P}(x_{1})C_{P}(x_{2})\dots C_{P}(x_{n})$ .

For general group actions

Suppose $p,q$ are distinct primes. Let $P$ be a finite $p$ -group, and $Q$ be an abelian $q$ -group that is not cyclic. Suppose $Q$ acts on $P$ by automorphisms. Suppose $x_{1},x_{2},\dots ,x_{n}$ are the non-identity elements of $Q$ , enumerated in any arbitrary order. Then, we have:

$P=C_{P}(x_{1})C_{P}(x_{2})\dots C_{P}(x_{n})$ .

Note that the version for general group actions follows directly from the version for faithful group actions. In fact, for an action that is not faithful, it suffices to take the products of centralizers of coset representatives of the kernel of the action.

Related facts

Particular cases

Centralizer product theorem for elementary abelian group

Corollaries/applications

Facts used

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

Given: Primes $p\neq q$ . A finite $p$ -group $P$ . An abelian non-cyclic subgroup $Q\leq \operatorname {Aut} (P)$ . $x_{1},x_{2},\dots ,x_{n}$ are the non-identity elements of $Q$ , enumerated in any order.

To prove: $P=C_{P}(x_{1})C_{P}(x_{2})\dots C_{P}(x_{n})$ .

Proof: We prove this by induction on the order of $P$ , combined with fact (1). Fact (1) settles the case for $P$ an elementary abelian $p$ -group.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Let $R=\Omega _{1}(Z(P))$ ; in other words, $R$ is the subgroup of $P$ comprising the identity element and elements of order $p$ in the center of $p$ . Then, $Q$ acts on $R$ by automorphisms, and $R$ is a nontrivial elementary abelian $p$ -group.	Fact (2)	$P$ is a finite $p$ -group		[SHOW MORE] This follows from the fact that $R$ is a characteristic subgroup of $P$ , so the action of $Q$ on $P$ restricts to $R$ . Also note that by fact (2), $R$ is nontrivial. Also by definition, $R$ is an elementary abelian $p$ -group.
2	There exists a non-identity $g\in Q$ such that $C_{R}(g)$ is nontrivial	Fact (1)	$Q$ is a finite abelian non-cyclic $q$ -group, $q\neq p$ .	Step (1)	[SHOW MORE] Note first that if the action of $Q$ on $R$ by automorphisms is not faithful, then there is a non-identity element $g\in Q$ such that $C_{R}(g)=R$ . On the other hand, if the action is faithful, then we can use fact (1) to conclude that $R$ is the product of $C_{R}(x)$ , $x$ ranging over the non-identity elements of $Q$ . In particular, there is some non-identity $g\in Q$ such that $C_{R}(g)$ is nontrivial.
3	With $g$ selected as in Step (2), $C_{R}(g)$ is a normal subgroup of $P$	Fact (3)		Steps (1), (2)	[SHOW MORE] Indeed, $R\leq Z(P)$ , so any subgroup of $R$ is central and hence normal in $P$ .
4	$Q$ stabilizes $C_{R}(g)$ and hence acts on $P/C_{R}(g)$		$Q$ is abelian		[SHOW MORE] Since $Q$ is abelian, $C_{R}(g)$ is $Q$ -invariant. We can see this as follows: suppose $h\in Q$ and $y\in C_{R}(g)$ . We need to show that $h\cdot y$ is also in $C_{R}(g)$ . For this, note that $g\cdot (h\cdot y)=(gh)\cdot y=(hg)\cdot y=h\cdot (g\cdot y)=h\cdot y$ , completing the proof (note the use of abelianness at the crucial step in between to interchange $g,h$ . Thus, the action of $Q$ on $P$ descends to an action on the quotient group.
5	Let ${\overline {P}}=P/C_{R}(g)$ . Then, ${\overline {P}}=\prod _{i=1}^{n}C_{\overline {P}}(x_{i})$		inductive hypothesis	Steps (3),(4)	[SHOW MORE] If $Q$ is faithful on ${\overline {P}}$ the induction hypothesis applies directly. Otherwise, let $Q_{1}$ be the kernel of the action and $\varphi :Q\to Q/Q_{1}$ be the quotient map. Consider the subset of $\varphi (x_{i})$ that are non-identity elements of $Q/Q_{1}$ . The induction hypothesis yields that ${\overline {P}}$ is the product of the centralizers of these. Taking inverse images, we obtain the desired result.
6	$P=\prod _{i=1}^{n}C_{P}(x_{i})$			Steps (2)-(5)	[SHOW MORE] From the previous step, the image of the product of $C_{P}(x_{i})$ under the quotient map to ${\overline {P}}$ is surjective. Also, one of the $x_{i}$ equals $g$ , so $C_{P}(x_{i})=C_{P}(g)\geq C_{R}(g)$ , and so the product of the centralizers contains the kernel of the quotient map to ${\overline {P}}$ . Hence, the product of the centralizers equals $P$ .