The Problem:
As the title asks: How many elements of order $5$ are there in a group of order $45$?
My Approach:
Let $G$ be a group of order $45$.
My first thought is to hit it with Sylow's Theorems, which tells me that $G$ has a normal subgroup of order $5$ (call it $N$) and a normal subgroup of order $9$. Clearly $4$ out of the $5$ elements in $N$ must be of order $5$ (since they must divide the order of $N$); and clearly none of these elements are in the normal subgroup of order $9$.
So, this means $G$ has at least $4$ elements of order $5$. Are there any more, though? The only thing I can think of is to investigate the possibility of a subgroup of order 15… but how would I know if $G$ has such a subgroup? And, if it did, how would I know whether it intersects $N$ trivially?
Best Answer
You already did 70% of the work:
Let $G$ be of order $45 = 5\cdot 3^2$. By Sylow's second theorem, the number of 5-sylow-groups $n_5$ must be either 1, 3, or 9. Out of those choices, only $n_5=1\equiv 1$ modulo 5. Therefore, only one subgroup of order 5 exists (normality isn't even important here).
Now note that every possible element of order five generates a subgroup of order five (in which it must be contained). I'm confident you can take it from here.
:-)