$a$ is of order $m$ if and only if $a^m = e$ and $a^k \neq e$ for all $0 < k < m$.
It is about order in Algebra. I sketch the proof. Is it correct? I need your help.
First ($\Rightarrow$) Let $| \langle a\rangle | = m$, then $\langle a\rangle = \{e, a, a^2, a^3, \cdots , a^{m-2}, a^{m-1} \}$ Since $| \langle a \rangle | = m$, $a^i$ is not equal to $a^j$ for $0 < i, j < m$ If $a^m$ is not equal to e, then $a^k = a^m$ for $0 < k < m$. because $| \langle a \rangle | = m$. So $a^{m-k} = e$. and it is contradiction that $| \langle a \rangle | = m$. therefore $a^m = e$. and If $a^k = e$ for $0 < k < m$. then $| \langle a \rangle |$ is not equal to $m$. So $a^k$ is not equal to $e$ for $0 < k < m$.
($\Leftarrow$) For $0 < i, j < m\; (i > j)$, if $a^i = a^j$, then $a^(i-j) = e$. But since $a^m = e$, $(i - j) = m$. It is contradiction to $0 < i, j < m$. Therefore $a^i$ is not equal to $a^j$ and number of $a^k$ for $0 < k < m$ is $(m - 1)$. So $| \langle a \rangle | = (m - 1) + 1 = m$ ($a^m = e$ is one).