Exercise from Artin's 2nd edition of Algebra.
How many elements of order $5$ might be contained in a group of order $20$?
My attempt:
By the third Sylow Theorem, $|Syl_{5} (G)| = 1, 2, 4$ and it is also $\equiv 1 \mod 5$, so it is 1. Thus theree is a unique subgroup of order $5$, which contains four elements of order $4$.
Correct?
Best Answer
Let $n_{5}$ denote the number of Sylow $5$-subgroups. By Sylow Theorems, $n_{5} \equiv 1 \mod 5$ and $n_{5}|4$ and both conditions imply that $n_{5} = 1$. Hence the number of non-identity elements of order $5$ is $n_{5}(5-1) = 4$.