If $G$ is a group, $a \in G$, and $k$ is a positive number, proof that $o(a^k) \le o(a)$.
First of all, I'm still confused on what $o(a^k)$ means.
I knew that if $\gcd(k, o(a))=1$ then $o(a^k)=o(a)$ is true by using this theorem $o(a^k)= \frac {n}{\gcd(n,k)}$, for $o(a)=n$.
So can I use the same theorem for $o(a^k) \le o(a)$?
If I can then does this simply can be proven with substituting $o(a^k)=\frac{n}{\gcd(n,k)}$ into $o(a^k) \le o(a)$ then I can have :
$$\frac{n}{\gcd(n,k)} \le n$$ $$\frac{1}{\gcd(n,k)}\le 1$$ for $$\gcd(n,k) \in \Bbb N$$