(related to Theorem: Order of Cyclic Group (Fermat's Little Theorem))

By hypothesis, \((G,\ast)\) is a cyclic group with a finite group order i.e. \(|G|=n < \infty,\), \(e\in G\) the neutral element and $a\in G.$

- From finite order of an element equals order of generated group it follows that the group $\langle a\rangle$ generated by by $a$ has a group order, which equals the order order of the element $\operatorname{ord}(a).$
- From Langrange's theorem it follows that $|\langle a\rangle|$ is a divisor of the group order $|G|.$

- Applying Langrange's theorem once again, we get
$$\begin{array}{rcl}a^{|G|}&=&a^{|\langle a\rangle|\cdot |G/\langle a\rangle|}\\
&=&(a^{|\langle a\rangle|})^{|G/\langle a\rangle|}\\
&=&(a^{\operatorname{ord}(a)})^{|G/\langle a\rangle|}\\
&=&e^{|G/\langle a\rangle|}\\
&=&e
\end{array}$$∎

**Modler, Florian; Kreh, Martin**: "Tutorium Algebra", Springer Spektrum, 2013