After using Sylow Theorems, how can we say how many elements of order 5 might be there in a group of order 20? [duplicate]

I know this question is asked here.

I get the first half which says that from the Sylow Theorems (3rd one), we can say that a group of 20, must have a unique subgroup which has order 5. But from here what is the reasoning behind elements having order 5?

Solution 1:

Each group of order $5$ is of the form $\mathbb Z/5 \mathbb Z$ because $5$ is prime. In the group $\mathbb Z/p \mathbb Z$ with $p$ prime you have $p-1$ generators, so there are $p-1$ elements of order $p$. That's because every non-identity generates the whole group since groups of prime order can't have proper non-trivial subgroups because such a subgroup's order would divide the group's order.

Edit: You don't even need to know that we are talking about $\mathbb Z/5 \mathbb Z$. Simply start with a group of prime order $p$ and follow the above argument to see that it has $p-1$ generators.