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

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Sylow $p$ -subgroups of $Aut ((Z / p)^{2})$ : $n_{p} = p + 1$

Example

For $E_{p} = (Z / p)^{2}$ , the Sylow $p$ -subgroups of $Aut (E_{p})$ have order $p$ and number $p + 1$ . Distinct ones meet trivially, and their union contains exactly $p^{2} - 1$ nonidentity elements. See For prime $p$ , $∣ Aut ((Z / p) \times (Z / p)) ∣ = (p^{2} - 1) (p^{2} - p)$ .

Facts & Assumptions

Given: The hypotheses and objects in the Example.

[L1]

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

[L2]

Let $G$ be a finite group, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $a \in N$ and $p ∤ m$ . A subgroup $P \leq G$ is a Sylow $p$ -subgroup when $∣ P ∣ = p^{a}$ . Equivalently, its order is the largest power of $p$ dividing $∣ G ∣$ . 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).

[L3]

For a finite group $G$ and a prime $p$ , let ${Syl}_{p} (G)$ be the set of Sylow $p$ -subgroups (def-sylow-p-subgroup). Define $n_{p} (G) : = ∣ {Syl}_{p} (G) ∣ .$ This cardinal is defined even before existence is proved because ${Syl}_{p} (G)$ 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).

[L4]

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

[L5]

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).

[L6]

Let $G$ be a finite group and $H \leq G$ . Then $∣ G ∣ = [G : H] ∣ H ∣ .$ Consequently, under the canonical embedding $ι : N \to Z$ , $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Verification

technique · direct

1.1L1L2L3L4L5L6givenalgebra

For $E_{p} = (Z / p)^{2}$ , the maps $u_{t} (x, y) = (x + t y, y)$ satisfy $u_{s} u_{t} = u_{s + t}$ , so $P = {u_{t} : t \in Z / p}$ has order $p$ . Because $∣ Aut (E_{p}) ∣ = p (p - 1)^{2} (p + 1)$ , it is a Sylow $p$ -subgroup.

2.1step 1.1givenalgebra

The common fixed subgroup of $P$ is $L = {(x, 0)}$ , since $u_{t} (x, y) = (x, y)$ for every $t$ exactly when $y = 0$ . A normalizer preserves $L$ ; conversely, a coordinate automorphism preserving $L$ conjugates each $u_{t}$ to another element of $P$ .

3.1step 2.1givenalgebra

Such an automorphism has the form $(x, y) \mapsto (a x + b y, d y)$ with $a, d \neq 0$ , giving $p (p - 1)^{2}$ choices. The normalizer-index formula therefore gives $n_{p} = p + 1$ .

4.1step 1.1step 3.1givenalgebra∎

Two distinct order- $p$ subgroups meet trivially, so their nonidentity elements are disjoint and number $(p + 1) (p - 1) = p^{2} - 1$ . At $p = 2$ , the normalizer has order $2$ in a group of order $6$ , giving three Sylow subgroups and three nonidentity elements. 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 pp-subgroups of Aut⁡((Z/p)2)Aut((Z/p)2): np=p+1np​=p+1