TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$

Statement

If $P$ is a nontrivial finite $p$ -group, then

$p ∣ ∣ Z (P) ∣ .$

In particular, the center $Z (P)$ contains a nonidentity element.

Facts & Assumptions

Given: A nontrivial finite $p$ -group $P$ .

[L1]

Nontriviality means $∣ P ∣ = p^{n}$ with $n \geq 1$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

For an action of $P$ on a finite set $X$ , $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L3]

Conjugation gives a homomorphism $P \to Aut (P)$ (The map $g \mapsto (x \mapsto g x g^{- 1})$ is a homomorphism $G \to Aut (G)$ with kernel $Z (G)$ and image $Inn (G)$ ).

[L4]

A homomorphism into a symmetric group defines an action (Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ ).

[L5]

The center is the set of elements commuting with every element of $P$ (The center $Z (G)$ of a group).

[L6]

Congruence modulo $p$ means divisibility of the difference by $p$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

Proof

technique · direct

1.1L3L4L5

By [L3] and [L4], $P$ acts on itself by conjugation. An element is fixed by all conjugations exactly when it lies in $Z (P)$ by [L5].

2.1step 1.1L2

Applying [L2] to this action gives $∣ P ∣ \equiv ∣ Z (P) ∣ (m o d p)$ .

3.1step 2.1L1L5L6algebra∎

By [L1], $p$ divides $∣ P ∣$ . Hence [L6] and step 2.1 show that $p$ divides $∣ Z (P) ∣$ . Since $Z (P)$ contains the identity and its cardinality is a positive multiple of $p > 1$ , it also contains a nonidentity element.

Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Every nontrivial finite pp-group has nontrivial center, in fact pp divides ∣Z(P)∣∣Z(P)∣

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$