Group of prime order is cyclic
Dependencies:
Let $G$ be a group where $|G|$ is prime.
Since $|G| > 1$, $G$ has an element $g$ which is not identity. $\operatorname{order}(g) > 1$, because $g$ is not identity.
$\operatorname{order}(g)$ divides $|G|$ and $|G|$ is prime. Therefore, $\operatorname{order}(g) = |G|$. This means $g$ is a generator of $G$.
Therefore, a group of prime order is cyclic and all non-identity elements are generators.
Dependency for: None
Info:
- Depth: 6
- Number of transitive dependencies: 15
Transitive dependencies:
- Group
- Coset
- Size of coset equals size of subset
- Identity of a group is unique
- Order of element in finite group is finite
- Subgroup
- Inverse of a group element is unique
- gH = H iff g in H
- Two cosets are either identical or disjoint
- Lagrange's Theorem
- Conditions for a subset to be a subgroup
- Cyclic Group
- Integer Division Theorem
- Order of cyclic subgroup is order of generator
- Order of element divides order of group