Finding the order of an element

Question

When we have permutation elements like $b=(12)(234)(1223)$ we can easily say that the order of each cycle is $2$, $3$ and $4$ respectively so the order of $b=\text{lcm}(2,3,4)$.

When we have $C_n$, does that mean the element with the highest order is $n$?

If this is the case then is it the same case for when wanting to find the highest order of elements in say $G=C_{5} \times C_{12} \times C_7$. So the highest order element is simply $\text{lcm}(5,12,7)=420$?

Answer 1

In $C_n\cong \mathbb{Z}_n$ the highest order for an element is $n$, that is the generator of $C_n$. When you have a product like $G=C_{a_1}\times C_{a_2}\times C_{a_3}\times \dots \times C_{a_m}$, highest order is, as you say, $\mathrm{lcm}(a_1,a_2,\dots, a_m)$, so you're right in that case.

One example of an element with this order may be the copy of each generator, $(1_{a_1},1_{a_2},\dots,1_{a_m})$.

Answer 2

The cyclic group indeed has order $n$ because it is the group generated by $\left<x\right>$ where $x$ has order $n$.

However, I think there is a problem with the product of groups because you have to show that it (the product) is isomorphic to a group of order $420$.