Let G be a finite group.
1) If $G$ is not abelian then $|Z(G)| \leq \frac{|G|}{4}$
2) If $|G:Z(G)|=n$ then each conjugacy class of $G$ contains at most $n$ elements.
So for 1, If G is not abelian then $Z(G)\neq G$, therefore $|Z(G)| < |G|$ and $|G| \geq4$.
Combining these two I managed to get the following:
$\frac{|G|}{|Z(G)|}\geq2$ and $\frac{|G|^{2}}{|Z(G)|}\geq2$
feels like I'm almost there, but not yet.
Regarding 2, I'll appreciate any guidance.
Thanks.
For $1$ , show that if $G$ is group such that $\frac{G}{Z(G)}$ is cyclic, then $G$ is abelian. Then, the fact that any group of prime order is cyclic tells you that $\frac{G}{Z(G)}$ cannot have order $2,3$ if it is non-trivial.
For the other one, let $G/Z(G) = \{p_1Z(G),p_2Z(G),...,p_nZ(G)\}$. Thus, for any element $d \in G$, we have $d \in p_iZ(G)$ for some $i$, then $d=p_ih$ for some $h \in Z(G)$, so that $p_i^{-1}d = h \in Z(G)$ for some $i$.
If $a \sim b$ then there is $c$ such that $a = cbc^{-1}$. Let $p_j$ be such that $p_j^{-1}c \in Z(G)$. Then, $p_j^{-1}ap_j = p_j^{-1}cbc^{-1}p_j = b$ since $p_j^{-1}c$ commutes with $b$.
Thus, we have shown that every element in the conjugacy class of $a$ is conjugate to $a$ via one of the $p_j$. Thus, the elements of $[a]$ are exactly the set $\{p_i^{-1}ap_i\}$. It is possible that some elements may coincide, but this has size at most $\frac{G}{Z(G)}$.
