Originally Posted by

**bulls6x** I will use the fact that the smallest non-abelian group has order 6, which is not hard to show.

Let $\displaystyle C(g_1),C(g_2),...,C(g_k) $ be the conjugacy classes of $\displaystyle G$. Assume WLOG that the first $\displaystyle m$ are non-central.

Suppose $\displaystyle |Z(G)|\not\le{\frac{n}{4}}.$ Then $\displaystyle |G/Z(G)|\lneq{4}$ Thus $\displaystyle |G/Z(G)|= 3,2, or 1. $ We will reach a contradiction for each case.

The case $\displaystyle |G/Z(G)|=1 $ is impossible since this would imply $\displaystyle G$ were abelian. So assume $\displaystyle |G/Z(G)|=2 $.

Let $\displaystyle g_i\in{G}$ with $\displaystyle i\le{m}$. Let $\displaystyle g\in{G\setminus{Z(G)}} $ with $\displaystyle g\neq{g_i}$. Consider $\displaystyle gg_{i}g^{-1} $

Since $\displaystyle g$ and $\displaystyle g_i$ are both noncentral and $\displaystyle G/Z(G) $ has order two we know that $\displaystyle gg_i\in{Z(G)}$

Thus $\displaystyle gg_ig^{-1} = g^{-1}gg_i=g_i$

Hence, we have shown that$\displaystyle gg_ig^{-1} = g_i$ for all $\displaystyle g\in{G}$ But this implies $\displaystyle g_i\in{Z(G)}$ which is a contradiction.

I don't have time to complete the last case. But it is fairly similar. Again use the fact that we know the group structure of $\displaystyle G/Z(G) $ and find the similar contradiction. Then use the class equation to get the inequality for $\displaystyle k$. I have not proven the case of equality but I'll leave that for someone else.