How many elements of order 5 may contain in a group of order 90? G - group of order 90
THE QUESTION:
How to found the count of elements of order 5 in this group?
 A: The elements of order $5$ are all in sylow p-subgroups.
How many are there? the number must be congruent to $1\bmod 5$ and divide $18$. so the possibilities are $1$ or $6$ .
Since the sylow p-subgroups do intersect trivially (lagrange) and each has $4$ of order $5$. The possibilites are $4$ and $24$.

Proof there can't be six $5$-sylow subgroups: since the group is of order $2$ by an odd number it has a subgroup of index $2$ (order $45$). Clearly such a subgroup is normal (Because it is of index $2$). 
Therefore our group is the semidirect product of a subgroup $H$ of order $45$ and $\mathbb Z_2$.
Groups of order $45$ are all abelian but all we need is the $5$-sylow is unique, which is straightforward since it must be $1\bmod 5$ and divide $9$.
So the elements of the group are of the form $(g,n)$. If $n$ is $1$ the order is even. So the elements of order $5$ are all of the form $(g,0)$ (notice $(g,0)^n=(g^n,0)$ since $0$ maps to the the trivial identity automorphism of $H$ under any homomorphism from $\mathbb Z_2$ to $Aut(H)$). Thus the elements of order $5$ are in bijective correspondance with those in $H$. Which are $4$.
