Solution: Let us first prove a general result.
For any $g \in G$ if $\text{ord}(g) = n $, then for any $k \in \mathbb{N} $, the order of $g^k$ will be $\frac{n}{\gcd (n, k)}$.
To prove the above, we need to show that
-
$\left( g^k \right) ^{\frac{n}{\gcd (n,k)}} = e$ and
-
if $\left( g^k \right) ^j = e$, then $\frac{n}{\gcd{n,k}} \mid j$.
Since $\text{ord}(g) = n$, the first point is clear.
\[
\left( g^k \right) ^{\frac{n}{\gcd (n,k)}} = \left( g^n \right) ^{\frac{k}{\gcd (n,k)}} = e.
\]
For the second point, we note that
\[
(g^k)^j = e \implies n \mid kj.
\]
So we have
\begin{align*}
n \mid kj \text{ and } n \mid nj & \implies n \mid \gcd (nj, kj) \\
& \implies n \mid \gcd (n,k)j \\
& \implies \frac{n}{\gcd (n,k)} \mid j.
\end{align*}
Now using the result that we have just proved we have
\begin{align*}
\text{ord} \left( g^6 \right) & = \frac{20}{\gcd (20,6)} = \frac{20}{2} = 10 \\[2ex]
\text{ord}\left( g^{13} \right) & = \frac{20}{\gcd (20, 13)} = \frac{20}{1} = 20.
\end{align*}