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