Find number of all $a \in G $ such that $o(a) =3$

Question

Let $G$ be a group and $|G|= 51$ find number of all $a \in G$ such that $o(a)=3$

My solution :

by this theorem : if $|G|=pq$ that $ p ,q$ are prime. If $ q\nmid p-1 $ then $\quad$ $G \cong \Bbb Z_{pq}$

So, since $51=3\times 17$ and $3 \nmid 16$ then $G \cong \Bbb Z_{51}$, so we have to find number of all $x \in \Bbb Z_{51}$ sucht that $\quad o(x)=3$. We claim that there are just two elements in $ \Bbb Z_{51}$ which have order of three.

Is this answer correct?

Answer 1

As $ \lvert G \rvert = 51 = 3 \cdot 17$, let $n_{3} = \lvert Syl_{3}(G) \rvert$. By the Sylow theorems we have that:

$n_{3} \vert 17$ and $n_{3} \equiv 1$ mod $3$

So as $17$ is prime, we have that either $n_{3} = 1$ or $n_{3} = 17$, but $ 17 \equiv -1$ mod $3$, so $n_{3} = 1$.

That means there is only one $3-Sylow$. As there is only one $3-Sylow$, it must be generated by one element, lets say $p$. Clearly $p$ has order three, and also $p^{2}$ has order three. It follows that there are two elements of order 3.