Group with an order? Let $G$ be a group and $g\in G$ with $\operatorname{ord}(g)=40$. Find the order of $a = g^8$, $b= g^5$, and what is $ab$?
I know how to solve the proff $\operatorname{ord}(a)=m$, $\operatorname{ord} (b) = n$, and $(m,n)=1$, then $\operatorname{ord} (ab) = mn$ which I think you can use here. 
How would you solve this example though? My TA did not do anything like this yet besides proofs. 
Can someone please show me how to solve this one example so I can see how to do it? 
 A: Any group generated by one element, like the group generated by $g$: $\langle g \rangle = \{e, g, g^1 \cdots, g^{39}\}$, is cyclic. Since we are given the order of $g$ is $40$ that means that $g^{40} = e$.
Note that $a^5 = (g^8)^5 = g^{40} = e, \;\; b^8 = (g^5)^8 = g^{40} = e$. 
So what are the orders of $a, b$?
Now, noting  $ab = g^8 g^5 = g^{13}$, and $\gcd(13, 40) = 1$, what must be the order of $ab$ be? Alternatively, since the group generated by $g$ is cyclic, you can use your proposition. If $m$ is the order of $a$, and $n$ is the order of $b$, and if $\gcd(m, n) = 1,$ then the order of $ab = mn$.
A: Hint: The least $n \geq 1$ such that $a^n = e$ is called the order of $a\in G$. You have been given that the order of $g$ is $40$. Now you need to apply above mentioned definition for $g^8$ and $g^5$.
A: $40 = 8 \cdot 5$ so $g^8$ has order 5 and $g^5$ order 8.
In general if $x$ has order $n$ and $dk = n$ then $x^d$ has order $k$ because
$n$ is the smallest such that $x^n = 1$ or $x^{dk} = (x^d)^k = 1$ is smallest.
