TITLE: Conditions (allegedly..) sufficient on a nonabelian finite group -- involving the number of conjugacy classes -- for the centre to be nontrivial
QUESTION [6 upvotes]: I recently encountered a -- may I say -- ''cute'' looking problem in a textbook on group theory. In my own notations, here is the hypothesis. For arbitrary group $F$ the (cardinal) number of conjugacy classes of $F$ is denoted by $\mathrm{c}(F)$ and for arbitrary subset $X \subseteq F$ the centraliser of $X$ in $F$ is written as $\mathrm{C}_{F}(X)$. For arbitrary set $M$ the full symmetric group on $M$ shall be denoted by $\Sigma(M)$. $\mathbf{Gr}$ refers to the category of groups.

Let $G$  be a finite nonabelian group and let us denote the largest prime divisor of the order $|G|$ by $p$. Given the relation $\mathrm{c}(G)>\frac{|G|}{p}$, prove that $\mathrm{Z}(G)$ is not trivial.

The problem follows the chapter which introduces group actions, representations by permutations, elementary properties of conjugation etc, so it should not require tools too sophisticated to solve (such as character theory).
My thoughts are to attempt Reductio ad absurdum by assuming the centre were trivial, to the effect that the number $\mathrm{c}(G)-1$ of nontrivial conjugacy classes is at least equal to $\frac{|G|}{p}$. This further means that at least one of these nontrivial classes -- say $C$ -- has cardinality strictly less than $p$, otherwise the union over all these classes -- expressible as $G \setminus \mathrm{Z}(G)=G \setminus \{1_G\}$ by virtue of our assumption and hence of cardinality $|G|-1$  -- has at the same time cardinality $\geqslant |G|$, which is absurd. Let us write $m=|C|$.
Thus, $G$ acts transitively (by conjugation) on $C$, action which induces a permutation representation $\rho \in \mathrm{Hom}_{\mathbf{Gr}}(G, \Sigma(C))$. The kernel of this representation is clearly given by $\mathrm{Ker} \rho=\mathrm{C}_G(C)$, so it follows that $(G:\mathrm{C}_G(C))|\ m!$. If it were the case that $p$ divides this index we would derive $p|m!$, which is equivalent to $p \leqslant m$ and contradicts one of the previous conclusions. Hence we gather that $p|\ |\mathrm{C}_G(C)|$.....and what of it?!
The reasoning above doesn't take into account the special property that $p$ has (of being the maximum among the set of prime divisors of $|G|$), which is indication that it is perhaps not the way to go searching for a proof...
Attempting to reason by contradiction by means of assuming the existence of a minimal (in the sense of order) counterexample also doesn't seem too promising, since the behaviour of either centres or conjugacy classes is rather hard to control when passing on to subgroups, in general.
All in all the problem doesn't seem to conceal anything too profound, but alas I see myself lacking in ideas to approach it. Any word of advice shall be greatly appreciated!

REPLY [2 votes]: Your statement is not true: take $G=S_3$, then the largest prime divisor of $|G|$ is $3$, and $c(S_3)=3 \gt \frac{6}{3}=2$, still $Z(S_3)=1$.
If $p$ is chosen to be the smallest prime, then it is true: assume that $Z(G)=1$, then note that for each $g \neq 1$, $\#Cl_G(g)=|G:C_G(g)| \geq p$. So the class formula gives us
$$|G| \geq 1 +p(c(G)-1)$$ and solving this for $c(G)$ we obtain $(1)$
$$c(G) \leq \frac{|G|+p-1}{p}.$$
But it is given that $$c(G) \gt \frac{|G|}{p}$$ whence $(2)$ $$ c(G) \geq \frac{|G|+p}{p}.$$
Combining $(1)$ and $(2)$ now leads to $p \leq p-1$ a contradiction.