Let G be a finite group and a from G.

Question

Let G be a finite group and a from G. Prove that $o(a)\le|G|$, where $o(a)$ is the order of elements, and $|G|$ group order.

Answer 1

$$o(a) = | \langle a \rangle| \le |G|.$$

Answer 2

Hint: Assume $o(a)>|G|$. Notice that $$\{e,a,a^2,\dots,a^{o(a)-1}\}$$ is a subgroup of $G$.

Answer 3

The map $f:\{0,\cdots, o(a)-1\}\to G$, $f(n)=a^n$ is injective. In fact, let $0\le m\le n<o(a)$ such that $a^n=a^m$. Then $0\le n-m <o(a)$ and $a^{n-m}=e$. By definition of $o(a)$, we have that $n-m=0$.