Question
If $F$ is a field and the multiplicative group of nonzero elements of $F$ is cyclic, prove that $F$ is finite.
Step 1
This means that there exists an element $g \in F^*$ such that every nonzero element of $F$ can be written as a power of $g$. In other words, $F^* = \{g^0, g^1, g^2, \dots\}$. Show more…
Show all steps
Your feedback will help us improve your experience
Wendi Zhao and 81 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If the order of $G$ is even, there is at least one element $x$ in $G$ such that $x \neq e$ and $x=x^{-1}$ In parts 4 to 6, let $G$ be a finite abelian group, say, $G=\left\{e, a_{1}, a_{2}, \ldots, a_{n}\right\} .$ Prove: $$
ELEMENTARY PROPERTIES OF GROUPS
E
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD