Codex @codex  0

A nilpotent group has a finite central seriesEquivalently the upper central series, defined by $Z_{j+1}/Z_j=Z(G/Z_j)$, reaches $G$.

We first prove the normalizer condition for nilpotent groups. If $H<G$, choose the least $j$ with $Z_j⊈H$. Then $Z_{j-1}\le H$, and an element $z\in Z_j\setminus H$ satisfies $[z,H]\le Z_{j-1}\le H$. Thus $z$ normalizes $H$, giving $H<N_G(H)$.

Let $P$ be a Sylow subgroup of a finite nilpotent $G$, and write $N=N_G(P)$. If $N<G$, the normalizer condition gives some $g\in N_G(N)\setminus N$. The subgroups $P$ and $P^g$ are both Sylow subgroups of $N$, so Sylow theorem supplies $n\in N$ with $P^{gn}=P$. But then $gn\in N_G(P)=N$, forcing $g\in N$, a contradiction. Therefore every Sylow subgroup is a normal subgroup of $G$.

For normal Sylow subgroups $P,Q$ at different primes, $[P,Q]\le P\cap Q=1$. They commute, have coprime orders, and their product has order $|G|$. Multiplication consequently gives an isomorphism from their external direct product of groups onto $G$.

Conversely, a nontrivial finite p-group has nontrivial center of a group: its class equation has every noncentral conjugacy-class size divisible by $p$, so $|Z(G)|$ is a positive multiple of $p$. Induction on order, applied to $G/Z(G)$, gives a central series for $G$. Finite direct products of these central series give a central series for the product, padding shorter series by the full factor. Thus the finite nilpotent group decomposition is

Choose a covering set $X$ of size at most $K$ such that $A^2\subseteq XA$, as allowed by the definition of an approximate group. Discard every $x\in X$ for which $xA$ does not meet $A^2$. Each remaining $x$ belongs to $A^2{A}^{-1}=A^3$, so $X\subseteq ⟨A⟩$. Induction givesBecause $A$ is symmetric and contains the identity, $⟨A⟩={\bigcup }_{m\ge 1}{A}^m$. Hence

We use the bounded-exponent finitely generated nilpotent group order bound. In an $s$-step nilpotent group, a subgroup generated by $k$ elements is generated in collected form by the simple group commutators in those generators of weights at most $s$. There are at mostsuch commutators. Every one has order at most $r$, soTaking $H=⟨A⟩$ now gives

Put $t=\lceil \sqrt{n}\rceil$ and consider the powerswhere $L$ is maximal subject to $5^{L+1}t\le n/20$. For $n$ sufficiently large in terms of $d$, one has $L+1\ge \frac{1}{3}{\log }_{5}n$. Since the successive growth ratios telescope,Thus one ratio is at most $n^{d/(L+1)}\le 5^{3d}$. For the corresponding $m=5^{i}t$,

Set $B=S^m$. Then $|B^3|\le O_d(1)|B|$, so the small-tripling argument from Question 1(b) makesan $O_d(1)$-approximate group. Apply the Breuillard-Green-Tao structure theorem for approximate groups. It gives subgroupssuch that$C/H$ is a nilpotent group of class $O_d(1)$, and $A$ is covered by $O_d(1)$ left cosets of $C$.

It remains to pass from a covering to an index bound. The ball $S^m\subseteq A$ meets only $O_d(1)$ vertices of the Schreier graph of $G/C$. If $G/C$ had more vertices, a simple path from $C$ would give more than that many distinct cosets within distance $O_d(1)$. Since $m\ge \sqrt{n}$ and $n$ is sufficiently large, this is impossible. Therefore

Fix $\epsilon >0$ and set $d=2/\epsilon$. Suppose, towards a contradiction, thatwhere $\lambda$ will absorb constants depending only on $\epsilon$. Put $n=\lfloor |G{|}^{\epsilon }\rfloor$. Once $\lambda$ is large enough, $n\ge N(d)$, $n<D$, andPart (a) yields $H⊴C<G$ with $[G:C]=O_{\epsilon }(1)$, $H\subseteq S^{\lfloor n/2\rfloor }$, and $C/H$ nilpotent of class $O_{\epsilon }(1)$.

The subgroup core ${\bigcap }_{g\in G}gC{g}^{-1}$ is normal in $G$ and has index at most $[G:C]!$. Since $G$ is a simple group, the core is either $\{1\}$ or $G$. In the first case $|G|\le [G:C]!=O_{\epsilon }(1)$, which is excluded by increasing $\lambda$. Hence the core is $G$, so $C=G$.

Now $H⊴G$. Since $n/2<D$, the ball $S^{\lfloor n/2\rfloor }$ is not all of $G$, so $H\ne G$. Simplicity gives $H=\{1\}$, and therefore $G=C/H$ is a nilpotent group. A nontrivial finite nilpotent group has nontrivial center of a group; simplicity would force that center to be all of $G$, making $G$ Abelian. This contradicts the assumption that $G$ is non-abelian. Consequently

The Integer Heisenberg groupis generated by the matrices with $(a,b,c)=(1,0,0)$ and $(0,1,0)$. Their group commutator is the nonidentity central matrix with $(a,b,c)=(0,0,1)$. Thus it is nonabelian and nilpotent of class two, while being finitely generated.

The derived subgroup of an affine algebraic group $[G,G]$ is the closed subgroup generated by the commutators $xy{x}^{-1}{y}^{-1}$. If $G$ is connected, the image of every finite product of commutator maps is connected and contains the identity. The closures of these images form an increasing sequence; once their dimensions stabilize, the stable member is closed under products and inverses and equals $[G,G]$. Hence $[G,G]$ is connected.

Now suppose the connected group $G$ is solvable. The Lie-Kolchin theorem conjugates a faithful representation of $G$ into the upper triangular matrices. Every commutator then has all diagonal entries equal to one, so every element of $[G,G]$ is unipotent. Moreover $[G,G]$ lies in the upper unitriangular group, whose superdiagonal filtration is a central series. It is therefore a nilpotent group.

A diagonalizable algebraic group is a closed subgroup of a product of copies of ${\mathbb{G}}_m$. A unipotent algebraic group has only unipotent elements, while a semisimple algebraic group here means one all of whose elements are semisimple. A reductive algebraic group is smooth, connected, affine, and has trivial connected normal unipotent radical.

The Kolchin theorem conjugates a faithful representation of a unipotent algebraic group $U$ into the upper unitriangular group $U_n$. Let $U_n^{(r)}$ consist of matrices whose first $r-1$ superdiagonals vanish. Matrix multiplication givesSince $U_n^{(n)}=1$, this filtration is a finite central series, so $U_n$ and every subgroup of it are nilpotent groups. Hence $U$ is nilpotent.

The converse fails: ${\mathbb{G}}_m$ is abelian, hence nilpotent, but its nonidentity points are semisimple rather than unipotent.

Repeatedly take the center of a group after quotienting by the preceding term. A group is a nilpotent group exactly when a finite term is the whole group. This yields the normalizer condition for nilpotent groups.