Element of Cyclic Group is not necessarily Generator
Jump to navigation
Jump to search
Theorem
Let $\gen g = G$ be a cyclic group.
Let $a \in G$
Then it is not necessarily the case that $a$ is also a generator of $G$.
Proof
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
Consider the subgroup $\gen 2$ of $\struct {\R_{\ne 0}, \times}$ generated by $2$.
By definition, $\gen 2$ is a cyclic group.
Consider the element $4 \in \struct {\R_{\ne 0}, \times}$.
We have that $4 = 2^2$.
Thus $4 \in \gen 2$.
There exists no $n \in \Z$ such that $4^n = 2$.
But $2 \in \gen 2$.
Thus $\gen 2$ is not generated by $4$.
That is, $4$ is not a generator of $\gen 2$.
Hence the result, by Proof by Counterexample.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 39$. Cyclic Groups