Let $G$ be a non-trivial finite group and let $p$ be the least prime divisor of $\vert G\vert $. If $k(G) > \frac{\vert G\vert}{p}$, prove that $Z(G) \neq 1$.
Suppose for a contradiction that $Z(G) = \lbrace 1\rbrace$. Then the Class Equation reads $\vert G\vert = 1 + \sum_{i=2}^{k} [G : C_G(x_i)]$. Where do I go from here in placing a bound on this? I cannot seem to arrive at a contradiction.
If $Z(G)=1$, every non-identity conjugacy class must have size $>1$ and hence $\geq p$. Thus the class equation gives $\# G\geq (k(G)-1)p+1$. But that is a little too much to ask for, since $k(G)-1\geq\frac{\# G}{p}$ (recall $k(G)$ is an integer and $p\mid \# G$)