any finite p-group is nilpotent

hbghlyj

A finite group $G$ is a p-group if its order is a power of a prime $p$.
The upper central series of a group $G$ is an ascending sequence of normal subgroups $Z_0(G) \subseteq Z_1(G) \subseteq Z_2(G) \subseteq \dots$ defined recursively:

• $Z_0(G) = \{e\}$
• $Z_1(G) = Z(G)$
• $Z_{i+1}(G)$ is the subgroup of $G$ such that $Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G))$.

A group $G$ is nilpotent if its upper central series terminates at $G$, meaning there exists an integer $c$ such that $Z_c(G) = G$.

The proof proceeds by induction on $n$, where $|G| = p^n$.

Lemma: The Center of a p-Group is Non-Trivial. For any non-trivial finite p-group $G$, its center $Z(G)$ is also non-trivial. This is proven using the class equation for $G$. Since $|G|$ is a power of the prime $p$, and the size of each conjugacy class must divide $|G|$, the class equation implies that $p$ must divide $|Z(G)|$.

The base case for the induction ($n=1$) is trivial, as a group of order $p$ is cyclic and thus abelian.
Assume that all p-groups of order $p^k$ for $k<n$ are nilpotent.
Inductive Step: Given a group $G$ of order $p^n$, we consider its center, $Z(G)$. By the lemma, $Z(G)$ is non-trivial. The quotient group $H=G/Z(G)$ is a p-group of smaller order, by the inductive hypothesis, $H$ is nilpotent. A standard theorem in group theory states that if $G/Z(G)$ is nilpotent, then $G$ must also be nilpotent. This is shown by relating the upper central series of $G/Z(G)$ to the upper central series of $G$. Specifically, if the upper central series of $G/Z(G)$ reaches $G/Z(G)$ in $c$ steps, the upper central series of $G$ reaches $G$ in at most $c+1$ steps.

• The fact that $H$ is nilpotent means that its upper central series terminates at $H$. Let this series be $\{e_H\} = Z_0(H) \subset Z_1(H) \subset \dots \subset Z_c(H) = H$.
• We now relate the upper central series of $G$ to that of $H$. Recall the definition $Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G))$.
  
  • For $i=0$, $Z_1(G)/Z_0(G) = Z(G/\{e\}) \implies Z_1(G) = Z(G)$.
  • By the Correspondence Theorem, there is a one-to-one correspondence between the subgroups of $H=G/Z(G)$ and the subgroups of $G$ that contain $Z(G)$. The upper central series of $H$ corresponds to a chain of subgroups in $G$, each containing $Z(G)$.
  • Let $\pi: G \to G/Z(G)$ be the canonical projection. Then $Z_1(H) = Z(G/Z(G))$. The pre-image $\pi^{-1}(Z_1(H))$ is a subgroup of $G$. By definition of the upper central series of $G$, we have $Z_2(G)/Z_1(G) = Z(G/Z_1(G)) = Z(G/Z(G))$. So, $Z_2(G)/Z(G) = Z_1(H)$. This implies $Z_2(G) = \pi^{-1}(Z_1(H))$.
  • In general, we have the relationship $Z_{i+1}(G)/Z(G) \cong Z_i(G/Z(G)) = Z_i(H)$ for $i \ge 1$.
• Since $H$ is nilpotent, there exists an integer $c$ such that $Z_c(H) = H$. Using the relationship from the previous step, this implies:$$Z_{c+1}(G)/Z(G) \cong Z_c(H) = H = G/Z(G)$$By the Correspondence Theorem, if the quotient groups are equal, the numerators must be equal. Therefore, $Z_{c+1}(G) = G$.
• We have shown that the upper central series of $G$ reaches $G$ in a finite number of steps. By definition, $G$ is nilpotent.

hbghlyj

For each $g \in G$ define a function $f_g: G \to G$ by $f_g(x)=[g, x]=g x g^{-1} x^{-1}$. This function is not a homomorphism, it's just a function. A group is nilpotent iff there exists some $n$ such that for all $g, f_g^n$ is the trivial map $G \to\{e\}$.

$G$ is nilpotent $\implies$ Each $f_g$ is nilpotent

This direction is true for any group $G$. We want to show that if $G$ is nilpotent, then there exists an integer $n$ such that $f_g^n(x) = e$ for all $g, x \in G$.

Proof:
Define the iterated commutator as $[g_1, g_2, \dots, g_k] = [g_1, [g_2, \dots, g_k]]$. Then $f_g^n(x) = [\underbrace{g, g, \dots, g}_{n \text{ times}}, x]$.

Since $G$ is nilpotent, its lower central series must terminate at the trivial subgroup. This means there exists some integer $c$ such that $\gamma_{c+1}(G) = \{e\}$.

We can show by induction that $f_g^k(x) \in \gamma_{k+1}(G)$ for all $g, x \in G$.
Base Case (k=1): $f_g^1(x) = [g, x]$. Since $g, x \in G = \gamma_1(G)$, their commutator $[g, x]$ is in $[G, G] = \gamma_2(G)$.
Inductive Step: Assume $f_g^k(x) \in \gamma_{k+1}(G)$. Now consider $f_g^{k+1}(x) = [g, f_g^k(x)]$. Here, $g \in G = \gamma_1(G)$ and by our hypothesis, $f_g^k(x) \in \gamma_{k+1}(G)$. By definition, their commutator must lie in $[G, \gamma_{k+1}(G)] = \gamma_{k+2}(G)$.

Thus, for any $k \ge 1$, $f_g^k(x) \in \gamma_{k+1}(G)$.

Since $G$ is nilpotent of class $c$, we know $\gamma_{c+1}(G) = \{e\}$. If we choose $n=c$, then for any $g, x \in G$, we have $f_g^c(x) \in \gamma_{c+1}(G) = \{e\}$. This implies $f_g^c(x) = e$.

This completes the proof for the forward direction. ✅

Each $f_g$ is nilpotent $\implies$ $G$ is nilpotent

This reverse implication is more subtle. A group where there exists an $n$ such that $[g, \dots, g, x] = e$ (with $g$ repeated $n$ times) for all $g, x$ is called an $n$-Engel group. The claim is that every $n$-Engel group is nilpotent.

General Case (Infinite Groups): This statement is false. There exist counterexamples of infinite groups that are $n$-Engel but not nilpotent (e.g., Golod-Shafarevich groups).

Finite Groups: The statement is true. If $G$ is a finite $n$-Engel group, then it must be nilpotent.

Proof (for finite groups):
This proof relies on a fundamental result from the theory of finite groups concerning Engel elements and the Fitting subgroup.

Definitions:

• An element $g \in G$ is a left Engel element if for every $x \in G$, there is an integer $n(x, g)$ such that $[\underbrace{g, \dots, g}_{n(x,g) \text{ times}}, x] = e$.
• The condition in the prompt states that there is a uniform integer $n$ that works for all $g$ and $x$. This means that every element of $G$ is a left Engel element.
• The Fitting subgroup, denoted $F(G)$, is the largest nilpotent normal subgroup of $G$. For a finite group, it is the product of all nilpotent normal subgroups.
• Key Theorem (Baer): For any finite group $G$, the set of all left Engel elements is precisely the Fitting subgroup $F(G)$.

The Argument:

• The condition given is that there exists an $n$ such that $f_g^n(x) = e$ for all $g, x \in G$.
• This directly implies that every element $g \in G$ is a left Engel element.
• According to Baer's theorem, the set of all left Engel elements is $F(G)$.
• Since every element of $G$ is a left Engel element, we must have $G = F(G)$.
• By definition, the Fitting subgroup $F(G)$ is nilpotent.
• Therefore, $G$ is nilpotent.