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Proof

(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 $a\in G.$

Proof of (1)

• 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 $|G|.$

Proof of (2)

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

Bibliography

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