I am currently trying to prove the following theorem.
For every $n\geq 1$ there exists $m\geq 1$ such that every finite group of order larger than $m$ contains more than $n$ conjugacy classes.
I have proved that, when $|G|=m$ with conjugacy classes $C_1,\cdots, C_n$ and $\ell_i=|C_{G}(x_i)|$ where $x_i\in C_i$ and where $C_G(x_i)$ is a centralizer of $x_i$ in $G$ with $\ell_1\geq \cdots \geq \ell_n$,
(a) $\frac{1}{\ell_1}+\cdots + \frac{1}{\ell_n}=1$
(b) For every $i$, $\ell_i\leq q_i(n)$ for some $q_i(n)$ which depends only on $n$.
It feels like I should use the $(a)$ and $(b)$ but I really have no idea. I am hoping to get at least a hint. Thanks in advance!
Lemma: For any $n$ and positive number $c$ there are finitely many positive integral solutions to $\frac{1}{\ell_1}+\frac{1}{\ell_2}+\dots+\frac{1}{\ell_n}= c$.
Proof: It's obvious for $n=1$ and then after that there are finitely many possibilities for the smallest $l_i$ (which we may assume w.l.o.g. is $l_1$) since we need $c= \frac{1}{\ell_1}+\frac{1}{\ell_2}+\dots+\frac{1}{\ell_n} < \frac{n}{l_i}$. Then we apply induction to these finitely many possibilities for $c -\frac{1}{l_1}$.
Okay so once we have this lemma, for each $n$ there is some maximum value $l_{max}$ among all of the $l_i$ values occurring in the finitely many solutions to $\frac{1}{\ell_1}+\frac{1}{\ell_2}+\dots+\frac{1}{\ell_n}= 1$. But in any group $G$ the identity is in its own conjugacy class, so one of the $l_i$ values you get must be $|G|$. This is a contradiction as soon as $|G| > l_{max}$.