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Equivalent forms of having a normal p complement

Statement

Let $G$ be a finite group and let $p$ be a prime. The following are equivalent.

(a) $G$ has a normal $p$ -complement. (b) $G$ has a normal $p^{'}$ -subgroup $K$ with $[G : K]$ a power of $p$ . (c) There is a Sylow $p$ -subgroup $Q$ of $G$ and an epimorphism $θ : G \to Q$ . (d) The product of any two $p^{'}$ -elements of $G$ is a $p^{'}$ -element. (e) Every $p^{'}$ -element of $G$ lies in $O_{p^{'}} (G)$ .

Moreover, if these conditions hold, then $O_{p^{'}} (G)$ is the normal $p$ -complement of $G$ , it is the unique normal $p$ -complement of $G$ , it equals the set of $p^{'}$ -elements of $G$ and the subgroup generated by them, and it is the kernel of every epimorphism $θ$ as in (c).

Facts & Assumptions

Given: A finite group $G$ and a prime $p$ , with $p^{'}$ -element terminology and the $p^{'}$ -core as in The p-prime core of a finite group and normal $p$ -complement as in Normal p complement and p nilpotent group.

[F1]

A normal $p$ -complement is a normal subgroup $K ⊴ G$ with $p ∤ ∣ K ∣$ and $[G : K]$ a power of $p$ ; for $P \in {Syl}_{p} (G)$ it is equivalent to $G = K P$ with $K \cap P = {1}$ , and $[G : K] = ∣ P ∣$ (Normal p complement and p nilpotent group, An internal semidirect product and a complement to a normal subgroup, Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[F2]

$O_{p^{'}} (G)$ is a normal $p^{'}$ -subgroup of $G$ containing every normal $p^{'}$ -subgroup of $G$ (The p-prime core of a finite group).

[F3]

$G$ has a Sylow $p$ -subgroup $P$ , of order $p^{a}$ where $∣ G ∣ = p^{a} m$ with $p ∤ m$ ; thus $∣ G ∣ = [G : P] ∣ P ∣$ and $p ∤ [G : P]$ (Sylow I: every finite group has a Sylow $p$ -subgroup, Sylow $p$ -subgroups of a finite group, Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[F4]

For $K ⊴ G$ the quotient $G / K$ is a group with $∣ G / K ∣ = [G : K]$ , and the natural map $G \to G / K$ is an epimorphism with kernel $K$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

[F5]

For a homomorphism $θ : G \to H$ one has $G / \ker θ ≅ im θ$ , $\ker θ ⊴ G$ and $im θ \leq H$ ; hence $∣ G ∣ = ∣ \ker θ ∣ \cdot ∣ im θ ∣$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ , The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

[F6]

If $K ⊴ G$ and $P \leq G$ then $P / (P \cap K) ≅ P K / K$ and $P K \leq G$ , so $∣ P K ∣ ∣ P \cap K ∣ = ∣ P ∣ ∣ K ∣$ (Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ , If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

[F7]

If $θ : G \to H$ is a homomorphism and $u \in G$ , then $θ (u)^{ord (u)} = e$ , so $ord (θ (u))$ divides $ord (u)$ , while $ord (v)$ divides $∣ H ∣$ for every $v$ in the finite group $H$ ; and $ord (v) = 1$ exactly when $v = e$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed, The order of every element of a finite group divides the order of the group, The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[F8]

If $H \leq G$ then $∣ H ∣$ divides $∣ G ∣$ , so a divisor of a power of $p$ that is prime to $p$ equals $1$ ; a divisor of a $p^{'}$ -number is a $p^{'}$ -number; and every integer greater than $1$ has a prime divisor (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ , Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime, A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[F9]

If the prime $p$ divides the order of a finite group $H$ , then $H$ has an element of order $p$ (Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ ).

[F10]

The subgroup generated by a set $Y$ consists of finite products of elements of $Y$ and their inverses, and conjugation is an automorphism of $G$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[F11]

$x^{n} = e$ if and only if $(x^{- 1})^{n} = e$ , so $ord (x^{- 1}) = ord (x)$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Proof

technique · direct

1.1

(a) implies (b) directly: a normal $p$ -complement is by definition a normal $p^{'}$ -subgroup of $p$ -power index.

F1given

1.2

(b) implies (c): let $K$ be a normal $p^{'}$ -subgroup with $[G : K] = p^{r}$ , so that $G / K$ is a finite group of order $p^{r}$ by [F4]. By [F6] applied to $K ⊴ G$ and a Sylow $P$ of $G$ given by [F3], $P / (P \cap K) ≅ P K / K \leq G / K$ ; here $∣ P \cap K ∣$ divides $∣ P ∣ = p^{a}$ and also divides $∣ K ∣$ , which is prime to $p$ , so $∣ P \cap K ∣ = 1$ by [F8]. Hence $∣ P K / K ∣ = ∣ P ∣ = p^{a}$ divides $∣ G / K ∣ = p^{r}$ , so $r \geq a$ . Since $∣ K ∣$ divides $∣ G ∣ = p^{a} m$ and is prime to $p$ , every prime divisor of $∣ K ∣$ divides $m$ , so $∣ K ∣$ divides $m$ and $m / ∣ K ∣$ is an integer prime to $p$ ; the identity $p^{r} = ∣ G ∣ / ∣ K ∣ = p^{a} (m / ∣ K ∣)$ therefore forces $r = a$ and $∣ K ∣ = m$ . Hence $∣ P K ∣ = ∣ P ∣ ∣ K ∣ / ∣ P \cap K ∣ = p^{a} ∣ K ∣ = p^{a} m = ∣ G ∣$ by [F6], so $P K = G$ .

F3F4F6F8algebra

1.3

(d) implies (e): let $x$ be a $p^{'}$ -element and let $Y$ be the set of conjugates of $x$ ; by (d) and [F11] every finite product of elements of $Y$ is a $p^{'}$ -element, so every element of $⟨ Y ⟩$ is a $p^{'}$ -element by [F10]. If a prime $ℓ$ divided $∣ ⟨ Y ⟩ ∣$ , then $⟨ Y ⟩$ would contain an element of order $ℓ$ by [F9], which forces $ℓ \neq p$ ; hence $p ∤ ∣ ⟨ Y ⟩ ∣$ , and $⟨ Y ⟩$ is a $p^{'}$ -subgroup. It is normal: $g ⟨ Y ⟩ g^{- 1} = ⟨ g Y g^{- 1} ⟩ = ⟨ Y ⟩$ for every $g \in G$ , by [F10] and the fact that $Y$ consists of all conjugates of $x$ . Hence $⟨ Y ⟩$ is a normal $p^{'}$ -subgroup and $⟨ Y ⟩ \leq O_{p^{'}} (G)$ by [F2], so $x \in O_{p^{'}} (G)$ .

F2F9F10F11assume-hyp

1.4

(e) implies (a): write $∣ G ∣ = p^{a} m$ with $p ∤ m$ as in [F3], and let $Q$ be a Sylow $q$ -subgroup of $G$ for each prime $q \neq p$ dividing $∣ G ∣$ , which exists by [F3] applied to $Q$ in place of $P$ (Sylow I: every finite group has a Sylow $p$ -subgroup). Every element of $Q$ has order dividing $∣ Q ∣ = q^{b}$ , hence prime to $p$ , so every element of $Q$ lies in $O_{p^{'}} (G)$ by (e); thus $Q \leq O_{p^{'}} (G)$ and $∣ Q ∣$ divides $∣ O_{p^{'}} (G) ∣$ by [F8]. Multiplying over the primes $q \neq p$ , the $p^{'}$ -number $m$ divides $∣ O_{p^{'}} (G) ∣$ ; since $O_{p^{'}} (G)$ is a $p^{'}$ -group by [F2], $∣ O_{p^{'}} (G) ∣$ divides $m$ , so $∣ O_{p^{'}} (G) ∣ = m$ .

F2F3F8assume-hyp

1.5

A subgroup is a $p^{'}$ -group exactly when all its elements are $p^{'}$ -elements; hence, by [F2] and (e), the set of $p^{'}$ -elements of $G$ is contained in $O_{p^{'}} (G)$ and contains all elements of $O_{p^{'}} (G)$ , so it equals $O_{p^{'}} (G)$ and generates it.

F2F7assume-hyp

2.1

Every $g \in G$ therefore has a unique expression $g = u k$ with $u \in P$ , $k \in K$ : existence is step 1.2, and uniqueness follows from $P \cap K = {1}$ , since $u k = u^{'} k^{'}$ gives $u^{' - 1} u = k^{'} k^{- 1} \in P \cap K$ . Defining $θ (g) : = u$ , one has $θ (g_{1} g_{2}) = θ (g_{1}) θ (g_{2})$ for all $g_{1}, g_{2}$ : writing $g_{i} = u_{i} k_{i}$ and using $k_{1} u_{2} = u_{2} (u_{2}^{- 1} k_{1} u_{2})$ with $u_{2}^{- 1} k_{1} u_{2} \in K$ , the product is $(u_{1} u_{2}) (u_{2}^{- 1} k_{1} u_{2} k_{2})$ . Thus $θ : G \to P$ is a homomorphism, it is onto because $θ (u) = u$ , and its kernel is $K$ .

F1F6step 1.2algebra

2.2

(c) implies (d): let $θ : G \to Q$ be an epimorphism onto a Sylow $Q$ of $G$ . By [F5], $∣ G ∣ = ∣ \ker θ ∣ \cdot ∣ Q ∣$ , so $∣ \ker θ ∣ = ∣ G ∣ / ∣ Q ∣ = [G : P]$ , a $p^{'}$ -number by [F3]. For a $p^{'}$ -element $x \in G$ , [F7] makes $ord (θ (x))$ a divisor of $ord (x)$ , hence prime to $p$ , and also a divisor of $∣ Q ∣ = p^{a}$ , hence a power of $p$ ; so $ord (θ (x)) = 1$ and $θ (x) = e$ by [F8] and [F7]. Thus every $p^{'}$ -element of $G$ lies in $\ker θ$ , and since $\ker θ$ is a $p^{'}$ -group, any product of two $p^{'}$ -elements lies in $\ker θ$ and has order dividing $∣ \ker θ ∣$ , hence is a $p^{'}$ -element by [F7] and [F8].

F3F5F7F8step 1.2

2.3

Consequently $∣ P \cap O_{p^{'}} (G) ∣ = 1$ : the order of the intersection divides both $p^{a}$ and the $p^{'}$ -number $m$ by [F8], so $∣ P O_{p^{'}} (G) ∣ = ∣ P ∣ ∣ O_{p^{'}} (G) ∣ = p^{a} m = ∣ G ∣$ by [F6], and $P O_{p^{'}} (G) = G$ . Since $O_{p^{'}} (G) ⊴ G$ by [F2] and $[G : O_{p^{'}} (G)] = ∣ G ∣ / m = p^{a}$ is a power of $p$ , $O_{p^{'}} (G)$ is a normal $p$ -complement of $G$ : (a) holds.

F1F2F6F8step 1.4

2.4

The moreover clauses. Assume (a)–(e) hold. Any normal $p$ -complement $K$ is a normal $p^{'}$ -subgroup, so $K \leq O_{p^{'}} (G)$ by [F2]; and $∣ K ∣ = ∣ G ∣ / [G : K] = m = ∣ O_{p^{'}} (G) ∣$ by [F1], [F3] and [F8], because $[G : K]$ is a $p$ -power dividing $∣ G ∣ = p^{a} m$ and $∣ K ∣$ is a $p^{'}$ -number. Hence $K = O_{p^{'}} (G)$ , so $O_{p^{'}} (G)$ is the unique normal $p$ -complement.

F1F2F3F8step 1.4

3.1

Finally let $θ : G \to Q$ be an epimorphism as in (c). By step 2.2 every $p^{'}$ -element lies in $\ker θ$ , while $∣ \ker θ ∣ = [G : P] = m$ is a $p^{'}$ -number; so $\ker θ$ is a normal $p^{'}$ -subgroup with $[G : \ker θ] = ∣ P ∣$ a $p$ -power, hence a normal $p$ -complement, hence equal to $O_{p^{'}} (G)$ by step 2.4. ∎

F1F5step 2.2step 2.4

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93 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Paul Flavell, An Introduction to Transfer and Fusion in Finite Groups, §§2–5 (standard reference, not scraped)