The finite Heisenberg group is the unique Sylow $p$-subgroup of its coordinate upper-triangular group

Given: The hypotheses and objects in the Example.

Let $G$ and $H$ be groups. Their external direct product has underlying set $G\times H:=\{(g,h):g\in G,h\in H\}$ and componentwise operation $(g,h)(g^{\mathrm{\prime }},h^{\mathrm{\prime }}):=(g{g}^{\mathrm{\prime }},h{h}^{\mathrm{\prime }})\mathrm{.}$ The fact that this operation makes $G\times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G\times H$ with componentwise multiplication).

For groups $G$ and $H$, the componentwise operation of def-external-direct-product-of-groups makes $G\times H$ a group. Its identity is $(e_G,e_H)$, and $(g,h{)}^{-1}=(g^{-1},h^{-1})\mathrm{.}$ Moreover the coordinate maps ${\pi }_G(g,h)=g$ and ${\pi }_H(g,h)=h$ are group homomorphisms. ($G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections).

An action of a group $H$ on a group $N$ by automorphisms is a homomorphism $\alpha :H⟶\mathrm{Aut}(N)\mathrm{.}$ Here automorphisms are those of def-group-isomorphism-and-automorphism. Writing ${\alpha }_h=\alpha (h)$, this means that every ${\alpha }_h$ is an automorphism of $N$, ${\alpha }_{hk}={\alpha }_h\circ {\alpha }_k$, and ${\alpha }_1={\mathrm{id}}_N$. Equivalently, by thm-group-actions-correspond-to-homomorphisms, it is a group action (def-group-action) on the underlying set of $N$ for which every acting permutation is an automorphism. (An action of a group $H$ on a group $N$ by automorphisms).

For an action $\alpha :H\to \mathrm{Aut}(N)$, the external semidirect product $N{⋊}_{\alpha }H$ is $N\times H$ with multiplication $(n,h)(n^{\mathrm{\prime }},h^{\mathrm{\prime }})=(n{\alpha }_h(n^{\mathrm{\prime }}),h{h}^{\mathrm{\prime }})\mathrm{.}$ ( The external semidirect product $N{⋊}_{\alpha }H$).

Let $\alpha :H\to \mathrm{Aut}(N)$ be an action by automorphisms. The multiplication $(n,h)(n^{\mathrm{\prime }},h^{\mathrm{\prime }})=(n{\alpha }_h(n^{\mathrm{\prime }}),h{h}^{\mathrm{\prime }})$ makes $N\times H$ a group with identity $(1_N,1_H)$ and inverse $(n,h{)}^{-1}=({\alpha }_{h^{-1}}(n^{-1}),h^{-1})\mathrm{.}$. ( The semidirect-product multiplication makes $N\times H$ a group).

In $N{⋊}_{\alpha }H$, the canonical copies $\bar{N}=\{(n,1):n\in N\}$ and $\bar{H}=\{(1,h):h\in H\}$ are subgroups, $\bar{N}$ is normal, their intersection is trivial, every element has a unique factorization $(n,1)(1,h)$, and $(1,h)(n,1)(1,h{)}^{-1}=({\alpha }_h(n),1)\mathrm{.}$ (The canonical copy of $N$ is normal, the canonical copy of $H$ is a complement, and conjugation induces the action).

A Sylow $p$-subgroup of a finite group is normal if and only if it is the unique Sylow $p$-subgroup. (A Sylow $p$-subgroup is normal if and only if it is unique).

For every prime $p$, the operations of addition and multiplication on $\mathbb{Z}\mathrm{/}p$ make it a field (def-field). (For every prime $p$, the two operations on $\mathbb{Z}\mathrm{/}p$ make it a field).

Let $n$ be a positive integer. Every class in $\mathbb{Z}\mathrm{/}n$ (def-integers-modulo-n) contains exactly one integer $r$ with $0\le r<n$. Consequently the map $r⟼[r{]}_n(0\le r<n)$ is a bijection from the von Neumann natural $n$ to $\mathbb{Z}\mathrm{/}n$, and $\mathrm{\mid }\mathbb{Z}\mathrm{/}n\mathrm{\mid }=n$. This includes $n=1$, where the only representative is $0$. For $n=0$, the map $a\mapsto [a{]}_0$ is a bijection $\mathbb{Z}\to \mathbb{Z}\mathrm{/}0$. (For $n\ge 1$, every class in $\mathbb{Z}\mathrm{/}n$ has one representative $r$ with $0\le r<n$, so $\mid \mathbb{Z}\mathrm{/}n\mid =n$; while $\mathbb{Z}\mathrm{/}0$ is in bijection with $\mathbb{Z}$).

For $n\ge 1$, the unit group is $(\mathbb{Z}\mathrm{/}n{)}^{\times }:=\{u\in \mathbb{Z}\mathrm{/}n:\text{some }v\in \mathbb{Z}\mathrm{/}n\text{ satisfies }uv=[1{]}_n\},$ and Euler's totient is $\phi (n):=\mathrm{\mid }(\mathbb{Z}\mathrm{/}n{)}^{\times }\mathrm{\mid }$. (The unit group $(\mathbb{Z}\mathrm{/}n{)}^{\times }$ and Euler's totient $\phi (n)=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid$ for $n\ge 1$).

Euler's totient satisfies $\phi (1)=1$. If $p$ is prime (def-prime), then $\phi (p)=p-1.$. ($\phi (1)=1$, and $\phi (p)=p-1$ for every prime $p$).

1. If $A$ and $B$ are finite then $A\times B$ is finite and $\mid A\times B\mid =\mid A\mid \cdot \mid B\mid$ (def-finite-cardinality). 2. Let $m\in \mathbb{N}$ and let $A_0,\dots ,A_{m-1}$ be finite sets. Write ${\prod }_{i<m}{A}_i:=\{f:f\text{ is a function with domain }m\text{ and }f(i)\in A_i\text{ for every }i<m\}\mathrm{.}$ Then ${\prod }_{i<m}{A}_i$ is finite and $\mid {\prod }_{i<m}{A}_i\mid ={\prod }_{i<m}\mid A_i\mid$, the right-hand product being the $\mathbb{N}$-valued one of def-nat-finite-sum-and-product. (The product rule: $\mid A\times B\mid =\mid A\mid \mid B\mid$, and $\mid {\prod }_{i<m}{A}_i\mid ={\prod }_{i<m}\mid A_i\mid$).