Question

If $\mathit{G}$ is a finite group of order $\mathit{n}\mathbf{}$and $\mathit{a}\mathbf{\in }\mathit{G}$, prove that $\left|a\right|\mathbf{\le }\mathit{n}$. [Hint: Consider the $\mathit{n}\mathbf{+}\mathbf{1}$ elements $\mathit{e}\mathbf{=}{\mathit{a}}^{\mathbf{0}}\mathit{a}\mathbf{,}{\mathit{a}}^{\mathbf{2}}\mathbf{,}{\mathit{a}}^{\mathbf{3}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{a}}^{\mathbf{n}}$. Are they all distinct?]. Thus every element in a finite group has finite order. The converse, however, is false; see Exercise 25 in Section 8.3 for an infinite group in which every element has finite order.

Short answer

It is proved that $\left|a\right|\le n$.

Step-by-step solution

Determine given functions

Consider that $G$ is a finite group of order $n$and $a\in G$.

As $G$has $n$elements, the list of $n+1$ elements $e=a^0,a^1,...,a^n$ should have at least one repetition, that is, there must be $0\le i\le j\le n$, so that $a^i=a^j$.

Determine |a|≤n

Now, using Theorem 7.2 statement (2), it is clear that $a$ has finite order and it is known as $k$.

By Theorem 7.9 (2), it is clear that$i=j\left(\mathrm{mod}k\right)$ and $k$divides $0<j-i\le n$.

That means, $k\le j-1\le n$ as required.

Hence, $\left|a\right|\le n$.