ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-17

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Sylow subgroups of $Aff (Z / 5)$ : $n_{2} = 5$ and $n_{5} = 1$

Example

In $Aff (Z / 5)$ , the translation subgroup is the unique Sylow $5$ -subgroup and the five point stabilizers are the Sylow $2$ -subgroups. Thus $n_{5} = 1$ and $n_{2} = 5$ . See Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$ .

Facts & Assumptions

Given: The hypotheses and objects in the Example.

[L1]

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

[L2]

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).

[L3]

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

[L4]

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

Verification

technique · direct

1.1L1L2L3L4givenalgebra

Composition identifies the affine maps $x \mapsto a x + b$ , with $a \in F_{5}^{\times}$ and $b \in F_{5}$ , with $F_{5} ⋊ F_{5}^{\times}$ . The multiplier map has the translation subgroup as its kernel, so that normal subgroup has order $5$ and is the unique Sylow $5$ -subgroup.

2.1step 1.1givenalgebra

For each $c \in F_{5}$ , the stabilizer of $c$ consists of the four maps $x \mapsto a (x - c) + c$ . It has order $4$ , the full $2$ -part of the group order $20$ , and hence is Sylow.

3.1step 2.1givenalgebra∎

The five point stabilizers are distinct, and Sylow III permits at most five Sylow $2$ -subgroups. Consequently they are all of them, so $n_{2} = 5$ and $n_{5} = 1$ . This proves the stated claim.

Depends on

Used by

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Dependency tree · two levels

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Sources

Keith Conrad, Consequences of the Sylow Theorems, Sections 1-5 (standard reference, not scraped)

Sylow subgroups of Aff⁡(Z/5)Aff(Z/5): n2=5n2​=5 and n5=1n5​=1