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Cyclic sylow does not alone imply a normal p complement

Statement refuted

Let $A_{5}$ be the alternating group on the five letters $0, 1, 2, 3, 4$ and let $p = 5$ (The alternating group $A_{n} = \ker (sgn)$ of even permutations, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation). Then:

$∣ A_{5} ∣ = 60$ and every Sylow $5$ -subgroup $P$ of $A_{5}$ is cyclic of order $5$ (Sylow $p$ -subgroups of a finite group, A finite group of prime order is cyclic and every nonidentity element generates it);
$A_{5}$ has no normal $5$ -complement (Normal p complement and p nilpotent group);
for such a Sylow $P$ the automizer is $N_{A_{5}} (P) / C_{A_{5}} (P) ≅ C_{2}$ , so it is not a $5$ -group and the automizer criterion of Frobenius automizer criterion for p nilpotence fails.

Thus a cyclic Sylow $p$ -subgroup does not by itself imply the existence of a normal $p$ -complement: the defect is detected by the automizer, not by the isomorphism type of the Sylow subgroup.

Facts & Assumptions

Given: The alternating group $A_{5}$ and the prime $p = 5$ .

[F1]

$A_{5}$ has order $60$ : indeed $∣ S_{5} ∣ = 120$ and $2 ∣ A_{5} ∣ = 5!$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ , The alternating group $A_{n} = \ker (sgn)$ of even permutations).

[F2]

$A_{5}$ is simple, and $A_{5} \neq {1}$ because $∣ A_{5} ∣ = 60$ ( $A_{n}$ is simple for every $n \geq 5$ , [F1]).

[F3]

Sylow counting: $n_{5} (A_{5}) \equiv 1 (m o d 5)$ and $n_{5} (A_{5}) ∣ 12$ , so $n_{5} (A_{5}) \in {1, 6}$ because the divisors of $12$ are $1, 2, 3, 4, 6, 12$ ; and if $n_{5} (A_{5}) = 1$ then the unique Sylow $5$ -subgroup is fixed by conjugation by every element of $A_{5}$ , hence is normal, of order $5$ , so nontrivial and proper, contradicting [F2]; therefore $n_{5} (A_{5}) = 6$ and, whenever $P \in {Syl}_{5} (A_{5})$ , $[A_{5} : N_{A_{5}} (P)] = 6$ so $∣ N_{A_{5}} (P) ∣ = 10$ (Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$ , Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class, Sylow III*: $n_{p} (G) = [G : N_{G} (P)]$ , The number $n_{p} (G)$ of Sylow $p$ -subgroups, Normal subgroup: invariance under conjugation, Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , [F1], [F2]).

[F4]

Let $P \in {Syl}_{5} (A_{5})$ . Then $∣ P ∣ = 5$ by [F3] and [F1], so $P$ is cyclic (A finite group of prime order is cyclic and every nonidentity element generates it), and in particular abelian; a subgroup of order $5$ is generated by any of its nonidentity elements ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian, The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, 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, Subgroup, One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

[F5]

Automorphism group of a cyclic group of prime order: $Aut (P) ≅ (Z / 5 Z)^{\times}$ has order $4$ ( $Aut (C_{n}) ≅ (Z / n Z)^{\times}$ , Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[F6]

Conjugation action of the normalizer: for $P \leq G$ , the map $α : N_{G} (P) \to Aut (P)$ , $α (n) (q) = n q n^{- 1}$ , is a group homomorphism, because conjugation is an automorphism of $P$ for $n \in N_{G} (P)$ and composition of conjugations is conjugation by the product; its kernel is ${n \in N_{G} (P) : n q n^{- 1} = q$ for all $q \in P} = N_{G} (P) \cap C_{G} (P) = C_{G} (P)$ ; by the first isomorphism theorem $N_{G} (P) / C_{G} (P)$ is isomorphic to a subgroup of $Aut (P)$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup, The centralizer $C_{G} (H)$ of a subgroup, $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ , Conjugation $x \mapsto g x g^{- 1}$ is an automorphism, Monoid homomorphism and group homomorphism, The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, First isomorphism theorem for groups: $G / \ker f ≅ im f$ , 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 $∣ G ∣ / ∣ N ∣$ , Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential).

[F7]

Cycle signs and conjugations: a cycle of length $k$ has sign $(- 1)^{k - 1}$ , and the sign is a homomorphism, so a $5$ -cycle and a product of two transpositions are even and lie in $A_{5} = \ker (sgn)$ ; and for a cycle $σ = (a_{0} a_{1} a_{2} a_{3} a_{4})$ and a permutation $g$ one has $g σ g^{- 1} = (g (a_{0}) g (a_{1}) g (a_{2}) g (a_{3}) g (a_{4}))$ (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles, The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ , The alternating group $A_{n} = \ker (sgn)$ of even permutations, Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F8]

Automizer criterion: for a finite group $G$ , a prime $p$ and $P \in {Syl}_{p} (G)$ , the group $G$ has a normal $p$ -complement if and only if $N_{G} (Q) / C_{G} (Q)$ is a $p$ -group for every $Q$ with $1 \neq Q \leq P$ (Frobenius automizer criterion for p nilpotence, Normal p complement and p nilpotent group).

Counterexample

technique · direct

1.1

By [F1] and [F3], $∣ A_{5} ∣ = 60 = 5 \cdot 12$ and the exact power of $5$ dividing $∣ A_{5} ∣$ is $5$ ; so a Sylow $5$ -subgroup has order $5$ , and a group of order $5$ is cyclic by [F4]. Moreover $n_{5} (A_{5}) = 6$ . This is assertion 1.

F1F3F4

2.1

Put $σ : = (0 1 2 3 4)$ and $P : = ⟨ σ ⟩ \leq A_{5}$ . By [F7] the $5$ -cycle $σ$ is even, so $σ \in A_{5}$ ; $P$ has order $5$ because $σ$ has order $5$ ; hence $P \in {Syl}_{5} (A_{5})$ by step 1.1, and $P = {id, σ, σ^{2}, σ^{3}, σ^{4}}$ is abelian.

F4F7step 1.1

3.1

Put $t : = (1 4) (2 3)$ . By [F7] the sign of each transposition is $- 1$ , so $t$ has sign $(- 1)^{2} = 1$ and $t \in A_{5}$ . By the conjugation formula of [F7], $t σ t^{- 1} = (t (0) t (1) t (2) t (3) t (4)) = (0 4 3 2 1) = σ^{- 1}$ , which lies in $P$ ; hence $t \in N_{A_{5}} (P)$ .

F7step 2.1

4.1

Moreover $t \notin C_{A_{5}} (P)$ : if $t$ centralized $P$ then in particular $t σ t^{- 1} = σ$ , but step 3.1 gives $t σ t^{- 1} = σ^{- 1}$ , and $σ^{- 1} \neq σ$ because $σ$ has order $5 \neq 2$ so $σ^{2} \neq id$ , whereas $σ^{- 1} = σ$ would give $σ^{2} = id$ . Hence $N_{A_{5}} (P) \neq C_{A_{5}} (P)$ .

F4step 3.1

5.1

By [F6] the quotient $N_{A_{5}} (P) / C_{A_{5}} (P)$ is isomorphic to a subgroup of $Aut (P)$ , so its order divides $4$ by [F5]; it also divides $∣ N_{A_{5}} (P) ∣ = 10$ by [F3] and [F6]. Since it is not $1$ by step 4.1, its order divides $\gcd (4, 10) = 2$ and is $> 1$ ; hence $∣ N_{A_{5}} (P) / C_{A_{5}} (P) ∣ = 2$ , and this group is cyclic of order $2$ , that is $N_{A_{5}} (P) / C_{A_{5}} (P) ≅ C_{2}$ . This is assertion 3.

F3F5F6step 4.1

6.1

The group $C_{2}$ has order $2$ , which is not a power of the prime $5$ ; so the automizer of the nontrivial $5$ -subgroup $P \leq P$ is not a $5$ -group, and the criterion [F8] fails in its second condition; therefore $A_{5}$ has no normal $5$ -complement. This is assertion 2.

F4F8step 5.1

7.1

The same conclusion follows directly from simplicity: a normal $5$ -complement would be a normal subgroup of index $5$ , hence a proper nontrivial normal subgroup of $A_{5}$ , contradicting [F2].

F1F2step 6.1

8.1

Combining: the Sylow $5$ -subgroup $P$ is cyclic by assertion 1, yet $A_{5}$ has no normal $5$ -complement by steps 6.1 and 7.1, while its automizer $N_{A_{5}} (P) / C_{A_{5}} (P) ≅ C_{2}$ is not a $5$ -group by assertion 3. This is the announced counterexample, and it shows that cyclicity of the Sylow subgroup alone carries no normal $p$ -complement. ∎

step 1.1step 5.1step 6.1step 7.1

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Sources

David Craven, Finite Group Theory, Lecture 3 (standard reference, not scraped)
Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups, §§7.1–7.2 (standard reference, not scraped)