Center of $G$ is trivial and $p$ divides the order of $G$, show that $G$ has a non-trivial conjugacy class whose order is prime to $p$ 
Let $G$ be a group, and suppose the centre of $G$ is trivial. If $p$ divides $|G|$, show that $G$ has a non-trivial conjugacy class whose order is prime to $p$.

My attempted solution is below, it would be greatly appreciated if anyone can check whether it is correct! Thanks!
Because $Z(G)={e}$, there is only one conjugacy class with size $1$.
By orbit-stabilizer theorem, $|G| = |ccl(x)||CG(x)|$,
so either $p$ divides $ccl(x)$ or $p$ divides the centralizers of $x$.
If there exists some conjugacy classes with size divisible by $p$, let's group all of them together so its size becomes $np$ for some $n$. Since orbit partition, $np + 1 + |ccl(a_1)| + \dots = |G|$.
Since $p$ divides $|G|$ and $|ccl(a_i)| \neq 1$ there must exist at least one $|ccl(a_i)|$ with non-trivial size and whose order is prime to $p$.

I find it hard to read what you're writing since your formatting is poor but this is how I would do it (I think your reasoning also goes along those lines).
The class equation says that
$$ |G| = |Z(G)| + \sum_{i=1}^s \lvert\operatorname{Cl}(h_i)\rvert, $$
where $(h_i)_{i=1}^s$ is a set of representatives for the non-singleton conjugacy classes and $\operatorname{Cl}(h_i)$ is the conjugacy class of $h_i$. If $G$ had no non-trivial conjugacy class whose order is prime to $p$, then $p$ would divide all the $\lvert\operatorname{Cl}(h_i)\rvert$, and since here $|Z(G)| = 1$ and $p$ divides $|G|$, we would be forced to conclude that $p$ divides $1$, a contradiction.
