Estimate the Number of Conjugacy Classes of $G$ This is a series of questions in my book unanswered. Let $c(G)$ be the number of conjugacy classes in $G$. Define $\bar{c}(G):=\frac{c(G)}{|G|}$. Now we estimate the $\bar{c}(G)$ of a non-abelien $G$.
(a) $\bar{c}(G)\leq \frac{5}{8}$.
(b) There is a finite group $H$ with $\bar{c}(H)=\frac{5}{8}$.
(c) Suppose that there exists a prime number $p$ and an element $x\in G$ such that the cardinality of the conjugacy class of $x$ is divisible by $p$. Find a good/sharp upper bound for $\bar{c}(G)$.

I have no idea to solve these questions. Is there any special technologies to solve the kind question?
 A: Let $G$ be a non-abelian finite group. The class formula asserts that$|G|=|Z(G)| + \sum_i |Cl_G(x_i)|$, where $Cl_G(x_i)$ is the conjugacy class of certain non-central $x_i$. Now observe two things:
(1) $|G/Z(G)|$ cannot be $1, 2$ or $3$. Why? Because if $G/Z(G)$ is cyclic, then $G$ be must be abelian (group theory folklore!)
(2) $|Cl_G(x_i)|$ cannot be equal to $1$, because if it were then $x_i \in Z(G)$ and in the class formula we already have counted the contribution of the center of $G$.So what do we learn from these two observations? From (1): $|G/Z(G)| \geq 4$, so $|G| \geq 4|Z(G)|$. And from (2): $|Cl_G(x_i)| \geq 2$. 
Now let us combine these two facts into the class formula:
$|G| \geq |Z(G)| + (c(G)-|Z(G)|).2= 2c(G)-|Z(G)|\geq2c(G)-\frac{1}{4}|G|$. From this it follows that $c(G)\leq\frac{5}{8}|G|$.Can this bound be attained? Yes, have a look at $G=Q$, the quaternion group of order 8.For your question (c) try to do the same as above, but then you know that there is a conjugacy class $|Cl_G(x_i)| \geq p$. For the other conjugacy classes you still need to take $2$ as a minimal cardinality estimate.
Bonus remark 1. There is also a lower bound on the number of conjugacy classes $c(G)$. Note that if $x \in G-Z(G)$, then $Z(G) \subsetneq C_G(x)$, where $C_G(x)$ is the centralizer of $x$ in $G$. We can conclude that for non-central elements $x$, we have $|C_G(x)|\geq 2|Z(G)|$, or equivalently $|Cl_G(x)| \leq \frac{|G|}{2|Z(G)|}$. And hence, working with the class formula again 
$|G| \leq |Z(G)| + (c(G) - |Z(G)|).\frac{|G|}{2|Z(G)|} = |Z(G)| + \frac{|G|c(G)}{2|Z(G)|} - \frac{1}{2}|G|$. Hence $\frac{3}{2}|G| \leq |Z(G)| + \frac{|G|c(G)}{2|Z(G)|} \leq \frac{1}{4}|G| + \frac{|G|c(G)}{2|Z(G)|}$ and elaborating this a bit futher this yields the lower bound
$c(G) \geq 2\frac{1}{2}|Z(G)|$.
Note that also this bound is sharp, again take $G$ to be the quaternion group of order 8.
Bonus remark 2. Try to show that if $G$ is a non-abelian group of odd order, then even $c(G)\leq\frac{11}{27}|G|$. 
Bonus remark 3. If $G$ is a non-abelian $p$-group, then $c(G)\leq\frac{p^2+p-1}{p^3}|G|$. The bound is sharp by the way and is attained by so-called extra-special $p$-groups of order $p^3$.
