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The kernel of the automorphism action on $P / Φ (P)$ is a $p$ -group

Statement

Let $P$ be a finite $p$ -group. The kernel of

$ρ_{P} : Aut (P) ⟶ {Aut}_{F_{p}} (P / Φ (P))$

is a finite $p$ -group.

Facts & Assumptions

Given: A finite $p$ -group $P$ and $K : = \ker ρ_{P}$ .

[L1]

Every automorphism of a finite $p$ -group induces an $F_{p}$ -linear automorphism of $P / Φ (P)$ , and these form a homomorphism $ρ_{P}$ (Automorphisms act linearly on the Frattini quotient).

[L2]

If a $p^{'}$ -subgroup of $Aut (P)$ acts trivially on $P / Φ (P)$ , then it is trivial (Hall–Burnside: coprime automorphisms are detected on the Frattini quotient).

[L3]

If a prime $q$ divides the order of a finite group, that group contains an element of order $q$ (Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ ).

[L4]

A positive integer is the product of powers of its prime divisors, with uniquely determined exponents (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

Proof

technique · direct

1.1givenL1algebra

By [L1], $K$ is a subgroup of $Aut (P)$ . Since $P$ is finite, its automorphism group is a subgroup of the finite permutation group of its underlying set, so $K$ is finite.

2.1step 1.1L2L3L4algebra

If a prime $q \neq p$ divided $∣ K ∣$ , [L3] would give $α \in K$ of order $q$ . Then $⟨ α ⟩$ would be a nontrivial $p^{'}$ -subgroup acting trivially on the quotient, contradicting [L2].

3.1step 2.1L4algebra∎

By [L4], no prime other than $p$ occurs in $∣ K ∣$ , so $∣ K ∣$ is a power of $p$ . The exponent-zero case gives the trivial kernel, including $P = 1$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

M. van Beek, Topics in Finite p-Groups, Proposition 4.10 (standard reference, not scraped)

The kernel of the automorphism action on P/Φ(P)P/Φ(P) is a pp-group