# Centralizer product theorem for elementary abelian group

## Statement

Suppose $p,q$ are distinct primes. Suppose $P$ is an elementary abelian $p$-group, and $Q \le \operatorname{Aut}(P)$ is an abelian $q$-group that is not cyclic. Let $x_1, x_2, \dots, x_n$ be an enumeration of the non-identity elements of $Q$. Then, we have:

$P = C_P(x_1)C_P(x_2) \dots C_P(x_n)$.

where $C_P(x_i)$ denotes the set of fixed points in $P$ of the automorphism $x_i$.

## Related facts

### Generalizations

• Centralizer product theorem: The same result, except that $P$ is now an arbitrary $p$-group, rather than an elementary abelian group.

## Facts used

1. Maschke's averaging lemma

## Proof

Given: Primes $p \ne q$. An elementary abelian $p$-group $P$, a non-cyclic abelian $q$-subgroup $Q$ of $\operatorname{Aut}(P)$. $x_1, x_2, \dots, x_n$ are the non-identity elements of $Q$.

To prove: $P = C_P(x_1)C_P(x_2) \dots C_P(x_n)$.

Proof:

1. By fact (1), viewing $P$ as a vector space over the field of $p$ elements and $Q$ as a group acting on this vector space, we can decompose $P$ as a direct sum of irreducible $Q$-modules, say $P_1, P_2, \dots, P_r$.
2. Let $Q_i$ be the pointwise stabilizer of $P_i$ in $Q$. Then, $Q/Q_i$ is cyclic: $Q/Q_i$ is an abelian group of automorphisms of the vector space $P_i$, and $P_i$ is irreducible. This forces $Q/Q_i$ to be cyclic.
3. Each $Q_i$ is a nontrivial subgroup of $Q$: This follows from the previous step and the fact that $Q$ isn't cyclic.
4. For each $i$, pick a non-identity element $x$ of $Q_i$. Then, $C_P(x)$ contains $P_i$: This follows from the definition of $Q_i$.
5. $P = \sum C_P(x_i)$: Each $P_i$ is contained in $C_P(x)$ for some non-identity element $x$. Thus, $P$ is the sum of all $C_P(x)$, $x$ ranging over the non-identity elements of $P$.

Reverting to multiplicative notation yields the result.

## References

### Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 69, Theorem 3.3, Chapter 3, Section 3 (Complete reducibility), More info