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The multiplicative group of a finite field is cyclic
Statement
The multiplicative group of every finite field is cyclic.
Facts & Assumptions
Given: A finite field .
A finite field has a finite underlying set (Finite fields and their order).
Every field is an integral domain, and its nonzero elements are its units (Every field is a commutative ring with ; it is an integral domain, and it is a commutative division ring).
Every finite subgroup of the unit group of an integral domain is cyclic (Every finite subgroup of the unit group of an integral domain is cyclic).
Proof
By [L2], is the unit group of the integral domain . It is finite by [L1].
Apply [L3] to the finite subgroup of itself to conclude that it is cyclic.
Depends on
Used by
- Every finite extension of a finite field is simple Corollary
- For every finite field F_q and every n≥1, a monic irreducible polynomial of degree n exists Corollary
- F₄=F₂[t]/(t²+t+1) with complete addition and multiplication tables Example
- F₉=F₃[t]/(t²+1) and a generator of its multiplicative group Example
- The residue class of t generates F₈^× Example
- A field with q elements is the splitting field of x^q-x over its prime subfield Proposition
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 53 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- K. Conrad, Finite Fields, Theorem 1.6 (standard reference, not scraped)