Prime power order implies nilpotent

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group of prime power order) must also satisfy the second group property (i.e., nilpotent group)
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Statement

Verbal statement

Any group of prime power order is nilpotent.

Related facts

Similar facts for groups of prime power order

• Prime power order implies not centerless: This result is the key ingredient used to prove that any group of prime power order is nilpotent.
• Prime power order implies center is normality-large: This is a stronger version of the fact that any group of prime power order is centerless.

Related facts for possibly infinite p-groups

• Locally finite Artinian p-group implies hypercentral
• p-group not implies nilpotent

Breakdown for Lie rings

• Prime power order not implies nilpotent for Lie rings

Facts used

1. Prime power order implies not centerless

Proof

We prove the statement by showing that it is possible to construct an upper central series for the group. The proof proceeds by induction on the order of the group. The base case for induction, namely the case of a group of prime order, is clear.

For the induction step, suppose the result is true for all groups whose order is $p^d,d<r$. We want to show that the result is true for $p^r$. Let $G$ be a group of order $p^r,r\ge 1$. Then, by fact (1), $Z(G)$ is nontrivial, and thus $G/Z(G)$ has order $p^d$ with $d<r$. Thus $G/Z(G)$ is nilpotent, so has an upper central series, and pulling this back gives an upper central series for $G$.