## Representatives for conjugacy classes

Posted: February 19, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Let $G$ be a finite group and $g_1, \cdots, g_r$ a system of representatives  for the conjugacy classes of $G$. Prove that if the $g_i$ pairwise commute, then $G$ is abelian.

Solution. We will denote by $C(g)$ the centralizer of $g \in G$ in $G.$ Suppose that $G$ is not abelian. Then $|G| > r$ and we may assume that $Z(G) = \{g_1, \cdots, g_s \},$ for some $1 \leq s < r.$ Since the $g_i$ pairwise commute, we have $g_i \in C(g_j),$ for all $i$ and $j,$ and hence $|C(g_j)| \geq r$ for all $j.$ Thus $\displaystyle [G: C(g_j)] \leq \frac {|G|}{r},$ for all $j.$ So by the class equation $\displaystyle |G| = s + \sum_{j = s + 1}^r[G: C(g_j)] \leq s + \frac {r - s}{r}|G|,$ which gives us the contradiction $|G| \leq r. \ \Box$