How is number of conjugacy class related to the order of a group? Let $c(G)$ denote the number of conjugacy classes of a group $G$. I have to show that 
$$\lim_{n \to \infty} \inf _{|G|=n}c(G)=\infty.$$ That is, I have to show that $\exists $ a function $f:\mathbb{N} \to \mathbb{N}$ such that if $c(g)\leq n, $ then $|G|\leq f(n).$ 
I know that if $G$ is a finite group such that any two non-identity elements in $G$ are conjugate, then $G$ is isomorphic to $\mathbb{Z}_2$. The proof is simple, infact: 

It can easily be shown that $G$ acts transitively by conjugation on $G \setminus \{e\}$. Then, by  orbit-stabilizer theorem, the claim follows. 

I wonder if similar idea could be useful somehow. But I can't think of any to prove the claim above. Please help!
 A: This is a result that goes back to Landau (1903). A proof is also given in the short article "A bound for the number of conjugacy classes in a group" by M. Newman. Here is the proof from that article, but the idea is pretty much the same as in all the other proofs.
Lemma: Let $x_1, \ldots, x_n$ be positive integers such that $1/x_1 + \cdots+ 1/x_n = 1$. Then  $\max x_i \leq n^{2^{n-1}}$.
Proof: Assume $x_1 \leq \cdots \leq x_n$. Now $x_1 \leq n$, as otherwise $1/x_i > 1/n$ for all $n$ and then $1/x_1 + \cdots + 1/x_n > 1$. For $1 \leq r \leq n-1$, we have 
$$\frac{1}{x_{r+1}} + \cdots + \frac{1}{x_{n}} = 1 - (\frac{1}{x_{1}} + \cdots + \frac{1}{x_{r}}) = \frac{y}{x_1x_2\cdots x_r}$$
where $y$ is a positive integer. Then
$$\frac{n-r}{x_{r+1}} \geq \frac{y}{x_1x_2 \cdots x_r} \geq \frac{1}{x_1x_2 \cdots x_r}$$
which implies $x_{r+1} \leq (n-r) x_1 \cdots x_r \leq n x_1 \cdots x_r$. Because $x_1 \leq n$, we see by induction that $x_r \leq n^{2^{r-1}}$ for all $1 \leq r \leq n$.
We apply the lemma to prove
Claim: Let $G$ be a finite group with $k$ conjugacy classes. Then
$$|G| \leq k^{2^{k-1}}$$ 
Proof: Suppose $G$ has conjugacy classes of orders $c_1, \ldots, c_k$. Then $c_1 + \cdots + c_k = |G|$, so
$$\frac{1}{|G|/c_1} + \cdots + \frac{1}{|G|/c_k} = 1$$
By applying the lemma, we see that $|G| \leq \max |G|/c_i \leq k^{2^{k-1}}$.
A: For $G$ non-abelian, take for instance $G=D_n$, the dihedral group of order $2n$. It is well-known that 
$$c(D_n)= \left\{ \begin{array}{rcl}
\frac{n+3}{2} & \mbox{for}
& n \text{ odd} \\ \frac{n+6}{2} & \mbox{for} & n \text{ even}
\end{array}\right.$$ This also implies the infinity of your limit.
