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A finite group of prime order is cyclic and every nonidentity element generates it
Statement
Let be a finite group such that the positive integer is prime. Then every has order , satisfies , and hence generates . In particular, is cyclic.
Facts & Assumptions
Given: A finite group with identity , with prime, and an element with .
A prime integer satisfies , and every positive divisor of is or (Prime and composite integers: is prime when and its only positive divisors are and ).
The natural is positive, equals exactly when , and its image in divides ; the embedding is injective and preserves order (The order of a finite group and the order of an element, with when no positive power of is the identity, The order of every element of a finite group divides the order of the group, The naturals embed in the integers).
If are finite and , then (A subset of a finite set is finite, with , and equality holds if and only if ).
If a finite set contains and , then some element of differs from : otherwise , whose cardinality is (The cardinality of a finite set).
Proof
The positive integer divides the prime , so it is or . It is not because , hence by injectivity of .
The subgroup has cardinality , so .
Thus every nonidentity element generates . Since by [F1], these two integers differ; injectivity in [L1] gives , and [F2] supplies a nonidentity element. Consequently is cyclic.
Depends on
- The order of every element of a finite group divides the order of the group
- Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$
- The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity
- If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$
- The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups
- $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian
- A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$, and equality holds if and only if $B = A$
- The naturals embed in the integers
- The cardinality $\lvert A\rvert$ of a finite set
Used by
- An extraspecial p-group is nilpotent of class exactly two and its derived subgroup has order p Corollary
- An extraspecial p-group is the product of two maximal abelian subgroups meeting in its centre Corollary
- Every group of order p², for prime p, is abelian Corollary
- The indecomposable finite abelian groups are exactly the nontrivial cyclic groups of prime-power order Corollary
- There are exactly two isomorphism classes of groups of order 105 Corollary
- 1→⟨ i⟩→ Q₈→ Q₈/⟨ i⟩→1 does not split, with nonabelian middle group Counterexample
- The commutator pairing of an extraspecial p-group relative to a chosen generator of its centre Definition
- A degree-three Galois extension of ℚ inside ℚ(ζ₇) Example
- Cyclic simple groups of prime order Example
- The ℤ-module ℤ/pᵏ has length k Example
- For primes p<q, nontrivial actions of Cₚ on C_q exist exactly when p∣(q-1) and are unique up to automorphisms Lemma
- Two elements of an extraspecial p-group with nontrivial commutator generate an extraspecial subgroup of order p³ Lemma
- Three equivalent descriptions of an extraspecial p-group Proposition
- Cyclic and alternating simple families Remark
- Cauchy's theorem for finite abelian groups Theorem
- Classification of groups of order pq for primes p<q Theorem
- Schur-Zassenhaus conjugacy when the kernel or quotient is solvable Theorem
- The Frattini quotient is the largest elementary abelian quotient of a finite p-group Theorem
- The unit group modulo n is the product of its odd-prime cyclic factors and its explicit 2-power factor Theorem
Dependency tree · two levels
49 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
- Thomas W. Judson, Abstract Algebra: Theory and Applications, Lagrange's Theorem (standard reference, not scraped)
- Thomas W. Judson, Abstract Algebra: Theory and Applications, §6.2: Lagrange's Theorem (standard reference, not scraped)
- Thomas W. Judson, Abstract Algebra: Theory and Applications, §4.1: Cyclic Subgroups (standard reference, not scraped)