Sylow's Theorems, $p$-Groups and Nilpotent Groups — Examples

Given: The hypotheses and objects in the Example.

For a finite group $G$ and a prime $p$, let ${\mathrm{Syl}}_p(G)$ be the set of Sylow $p$-subgroups (def-sylow-p-subgroup). Define $n_p(G):=\mathrm{\mid }{\mathrm{Syl}}_p(G)\mathrm{\mid }\mathrm{.}$ This cardinal is defined even before existence is proved because ${\mathrm{Syl}}_p(G)$ is a subset of the finite power set of $G$; thm-sylow-first-theorem later shows it is nonzero. (The number $n_p(G)$ of Sylow $p$-subgroups).

Let $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then the number of Sylow $p$-subgroups satisfies $n_p(G)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p),n_p(G)\mid m\mathrm{.}$. (Sylow III: $n_p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and $n_p\mid m$ when $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$).

Let $n\in \mathbb{N}$, so that $n=\{0,1,\dots ,n-1\}$ (def-natural-numbers). The symmetric group on $n$ letters is $S_n:=\mathrm{Sym}(n)=\mathrm{Sym}(\{0,1,\dots ,n-1\}),$ the group of all bijections of $n$ under composition (def-symmetric-group), with the composition convention. (The finite symmetric group $S_n$, one-line notation, and cycle notation).

For $\sigma ,\tau \in S_n$, there is a $g\in S_n$ with $\tau =g\sigma g^{-1}$ if and only if $\sigma$ and $\tau$ have the same cycle type, including their numbers of fixed points. (Two elements of $S_n$ are conjugate if and only if they have the same cycle type).

Given: The hypotheses and objects in the Example.

For a finite group $G$ and a prime $p$, let ${\mathrm{Syl}}_p(G)$ be the set of Sylow $p$-subgroups (def-sylow-p-subgroup). Define $n_p(G):=\mathrm{\mid }{\mathrm{Syl}}_p(G)\mathrm{\mid }\mathrm{.}$ This cardinal is defined even before existence is proved because ${\mathrm{Syl}}_p(G)$ is a subset of the finite power set of $G$; thm-sylow-first-theorem later shows it is nonzero. (The number $n_p(G)$ of Sylow $p$-subgroups).

Let $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then the number of Sylow $p$-subgroups satisfies $n_p(G)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p),n_p(G)\mid m\mathrm{.}$. (Sylow III: $n_p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and $n_p\mid m$ when $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$).

For $n\in \mathbb{N}$, the alternating group is the kernel of the sign homomorphism, $A_n:=\ker (\mathrm{sgn}:S_n\to \{+1,-1\})=\{\sigma \in S_n:\mathrm{sgn}(\sigma )=1\}\mathrm{.}$ Thus $A_n$ consists exactly of the even permutations. The subgroup and normality assertions implicit in the word “group” follow from thm-image-subgroup-and-kernel-normal. (The alternating group $A_n=\ker (\mathrm{sgn})$ of even permutations).

The conjugacy classes of $S_n$ are in bijection with the tuples of nonnegative integers $(c_1,\dots ,c_n)$ satisfying ${\sum }_{k=1}^{n}k{c}_k=n\mathrm{.}$ For $n=0$, the unique empty tuple indexes the identity class of $S_0$. (The conjugacy classes of $S_n$ are indexed by the tuples $(c_1,\dots ,c_n)$ with $\sum k{c}_k=n$).

Given: The hypotheses and objects in the Example.

Let $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then the number of Sylow $p$-subgroups satisfies $n_p(G)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p),n_p(G)\mid m\mathrm{.}$. (Sylow III: $n_p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and $n_p\mid m$ when $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$).

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).

Let $N$ and $H$ be groups (def-group), and let $\alpha :H\to \mathrm{Aut}(N)$ be an action by automorphisms (def-action-by-automorphisms). The external semidirect product $N{⋊}_{\alpha }H$ is the set $N\times H$ with multiplication. ( The external semidirect product $N{⋊}_{\alpha }H$).

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).

Given: The hypotheses and objects in the Example.

Let $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then the number of Sylow $p$-subgroups satisfies $n_p(G)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p),n_p(G)\mid m\mathrm{.}$. (Sylow III: $n_p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and $n_p\mid m$ when $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$).

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).

Let $N$ and $H$ be groups (def-group), and let $\alpha :H\to \mathrm{Aut}(N)$ be an action by automorphisms (def-action-by-automorphisms). The external semidirect product $N{⋊}_{\alpha }H$ is the set $N\times H$ with multiplication. ( The external semidirect product $N{⋊}_{\alpha }H$).

For every prime $p$ and natural $k\ge 1$, $\phi (p^k)=p^k-p^{k-1}\mathrm{.}$ Equivalently, among the $p^k$ standard classes modulo $p^k$, the nonunits are exactly those whose standard representatives are divisible by $p$. (For a prime $p$ and $k\ge 1$, $\phi (p^k)=p^k-p^{k-1}$).

Let $n\ge 1$ and $a\in \mathbb{Z}$. Then $[a{]}_n$ is a unit of $\mathbb{Z}\mathrm{/}n$ (def-unit-group-modulo-n-and-euler-totient) if and only if $\gcd (a,n)=1,$ that is, if and only if $a$ and $n$ are coprime (def-coprime). Consequently the condition $\gcd (a,n)=1$ depends only on the class $[a{]}_n$. (For $n\ge 1$, $[a{]}_n$ is a unit if and only if $\gcd (a,n)=1$).

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 $\mid {\prod }_{i<m}{A}_i\mid ={\prod }_{i<m}\mid A_i\mid$).

Given: The hypotheses and objects in the Example.

For every prime $p$, $\mathrm{\mid }\mathrm{Aut}((\mathbb{Z}\mathrm{/}p)\times (\mathbb{Z}\mathrm{/}p))\mathrm{\mid }=(p^2-1)(p^2-p)\mathrm{.}$. (For prime $p$, $\mathrm{\mid }\mathrm{Aut}((\mathbb{Z}\mathrm{/}p)\times (\mathbb{Z}\mathrm{/}p))\mathrm{\mid }=(p^2-1)(p^2-p)$).

Let $G$ be a finite group, let $p$ be prime, and write $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $a\in \mathbb{N}$ and $p\nmid m$. A subgroup $P\le G$ is a Sylow $p$-subgroup when $\mathrm{\mid }P\mathrm{\mid }=p^a$. Equivalently, its order is the largest power of $p$ dividing $\mathrm{\mid }G\mathrm{\mid }$. This is a property of a subgroup and does not presume that such a subgroup exists; existence is proved in thm-sylow-first-theorem. (Sylow $p$-subgroups of a finite group).

For a finite group $G$ and a prime $p$, let ${\mathrm{Syl}}_p(G)$ be the set of Sylow $p$-subgroups (def-sylow-p-subgroup). Define $n_p(G):=\mathrm{\mid }{\mathrm{Syl}}_p(G)\mathrm{\mid }\mathrm{.}$ This cardinal is defined even before existence is proved because ${\mathrm{Syl}}_p(G)$ is a subset of the finite power set of $G$; thm-sylow-first-theorem later shows it is nonzero. (The number $n_p(G)$ of Sylow $p$-subgroups).

If $P$ is a Sylow $p$-subgroup of a finite group $G$, then $n_p(G)=[G:N_G(P)]\mathrm{.}$. (Sylow III*: $n_p(G)=[G:N_G(P)]$).

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 $G$ be a finite group and $H\le G$. Then $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }\mathrm{.}$ Consequently, under the canonical embedding $\iota :\mathbb{N}\to \mathbb{Z}$, $\mathrm{\mid }H\mathrm{\mid }$ divides $\mathrm{\mid }G\mathrm{\mid }$. (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

Given: The hypotheses and objects in the Example.

Let $G$ be finite, let $p$ be prime, and write $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then $G$ has a subgroup of order $p^a$, hence a Sylow $p$-subgroup (def-sylow-p-subgroup). (Sylow I: every finite group has a Sylow $p$-subgroup).

Let $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then the number of Sylow $p$-subgroups satisfies $n_p(G)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p),n_p(G)\mid m\mathrm{.}$. (Sylow III: $n_p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and $n_p\mid m$ when $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$).

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).

Let $p<q$ be primes. - If $p\nmid (q-1)$, every group of order $pq$ is cyclic. - If $p\mid (q-1)$, there are exactly two isomorphism classes of groups of order $pq$: the cyclic group $C_{pq}$ and one nonabelian semidirect product $C_q⋊C_p$. (Classification of groups of order $pq$ for primes $p<q$).

Every finite $p$-group is nilpotent. The trivial group is included and has nilpotency class zero. (Every finite $p$-group is nilpotent).

Given: The hypotheses and objects in the Example.

Every group of order $30$ has normal Sylow $3$- and $5$-subgroups and is not simple. (Every group of order $30$ has normal Sylow $3$- and $5$-subgroups and is not simple).

Let $p<q$ be primes. - If $p\nmid (q-1)$, every group of order $pq$ is cyclic. - If $p\mid (q-1)$, there are exactly two isomorphism classes of groups of order $pq$: the cyclic group $C_{pq}$ and one nonabelian semidirect product $C_q⋊C_p$. (Classification of groups of order $pq$ for primes $p<q$).

Let $\alpha ,\beta :H\to \mathrm{Aut}(N)$ be actions. If $u\in \mathrm{Aut}(N)$ and $v\in \mathrm{Aut}(H)$ satisfy ${\beta }_{v(h)}=u{\alpha }_h{u}^{-1}(h\in H),$ then $N{⋊}_{\alpha }H\cong N{⋊}_{\beta }H$. (Actions changed by automorphisms of the kernel and complement give isomorphic semidirect products).

Let $N$ and $H$ be groups (def-group), and let $\alpha :H\to \mathrm{Aut}(N)$ be an action by automorphisms (def-action-by-automorphisms). The external semidirect product $N{⋊}_{\alpha }H$ is the set $N\times H$ with multiplication. ( The external semidirect product $N{⋊}_{\alpha }H$).

Given: The hypotheses and objects in the Example.

For a finite group $G$, the following are equivalent: $G$ is nilpotent; every Sylow subgroup is normal; $G$ is the internal direct product of its Sylow subgroups; and every maximal subgroup of $G$ is normal. (Sylow and maximal-subgroup characterizations of finite nilpotence).

For $n\in \mathbb{N}$, the alternating group is the kernel of the sign homomorphism, $A_n:=\ker (\mathrm{sgn}:S_n\to \{+1,-1\})=\{\sigma \in S_n:\mathrm{sgn}(\sigma )=1\}\mathrm{.}$ Thus $A_n$ consists exactly of the even permutations. The subgroup and normality assertions implicit in the word “group” follow from thm-image-subgroup-and-kernel-normal. (The alternating group $A_n=\ker (\mathrm{sgn})$ of even permutations).

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).

Given: The hypotheses and objects in the Example.

For a finite group $G$, the Fitting subgroup is $F(G):={\prod }_{p\mid \mathrm{\mid }G\mathrm{\mid }}{O}_p(G),$ the product of its $p$-cores (def-p-core-of-a-finite-group). The factors are normal, so their finite product is a normal subgroup and does not depend on the order of multiplication. For the trivial group the product is empty and equals $1$. (The Fitting subgroup $F(G)={\prod }_p{O}_p(G)$ of a finite group).

For a finite group $G$, the Frattini subgroup is $\mathrm{\Phi }(G):=\bigcap \{M\le G:M\text{ is maximal proper}\}\mathrm{.}$ If $G=1$, the family is empty and its intersection inside $G$ is $G$ itself. Thus $\mathrm{\Phi }(1)=1$. (The Frattini subgroup $\mathrm{\Phi }(G)$ as the intersection of the maximal subgroups of a finite group).

For every finite group $G$, $F(G)$ is nilpotent and normal, and every normal nilpotent subgroup of $G$ is contained in $F(G)$. (The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group).

Let $n\in \mathbb{N}$, so that $n=\{0,1,\dots ,n-1\}$ (def-natural-numbers). The symmetric group on $n$ letters is $S_n:=\mathrm{Sym}(n)=\mathrm{Sym}(\{0,1,\dots ,n-1\}),$ the group of all bijections of $n$ under composition (def-symmetric-group), with the composition convention. (The finite symmetric group $S_n$, one-line notation, and cycle notation).

Given: The hypotheses and objects in the Example.

For a finite group $G$ and a prime $p$, the $p$-core $O_p(G)$ is the subgroup generated by all normal $p$-subgroups of $G$. There are finitely many such subgroups, their product is normal, and repeated use of $\mathrm{\mid }AB\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\cap B\mathrm{\mid }$ shows that the product is again a $p$-group. It contains every normal $p$-subgroup, so it is the unique largest normal $p$-subgroup. (The $p$-core $O_p(G)$ as the largest normal $p$-subgroup).

For a finite group $G$, the Fitting subgroup is $F(G):={\prod }_{p\mid \mathrm{\mid }G\mathrm{\mid }}{O}_p(G),$ the product of its $p$-cores (def-p-core-of-a-finite-group). The factors are normal, so their finite product is a normal subgroup and does not depend on the order of multiplication. For the trivial group the product is empty and equals $1$. (The Fitting subgroup $F(G)={\prod }_p{O}_p(G)$ of a finite group).

For a finite group $G$, the Frattini subgroup is $\mathrm{\Phi }(G):=\bigcap \{M\le G:M\text{ is maximal proper}\}\mathrm{.}$ If $G=1$, the family is empty and its intersection inside $G$ is $G$ itself. Thus $\mathrm{\Phi }(1)=1$. (The Frattini subgroup $\mathrm{\Phi }(G)$ as the intersection of the maximal subgroups of a finite group).

For every finite group $G$, $F(G\mathrm{/}\mathrm{\Phi }(G))=F(G)\mathrm{/}\mathrm{\Phi }(G)\mathrm{.}$. ($F(G\mathrm{/}\mathrm{\Phi }(G))=F(G)\mathrm{/}\mathrm{\Phi }(G)$ for every finite group).

Let $n\in \mathbb{N}$, so that $n=\{0,1,\dots ,n-1\}$ (def-natural-numbers). The symmetric group on $n$ letters is $S_n:=\mathrm{Sym}(n)=\mathrm{Sym}(\{0,1,\dots ,n-1\}),$ the group of all bijections of $n$ under composition (def-symmetric-group), with the composition convention. (The finite symmetric group $S_n$, one-line notation, and cycle notation).