Let $G$ be a group of order $n > 1$. Show that no conjugacy class can have order
greater than $\frac{n}{2}$.
My approach: For any integer $n>1$ there is no a divisor $d$ such that $\frac{n}{2}<d<n$. Also, $d=n$ is not possible, since the conjugacy class of the identity has only one element.
Is this a correct approach? if not, how could you continue with the exercise?
Best Answer
The cardinality of the conjugacy class of $x\in G$ is the index of the centralizer $C_G(x)=\{g\in G:xg=gx\}$. Thus we'd have $$ n=[G:C_G(x)]|C_G(x)|>\frac{n}{2} |C_G(x)| $$ and so $|C_G(x)|<2$. Thus $C_G(x)$ is the trivial subgroup; since $x\in C_G(x)$ we conclude that $x=1$. But then $C_G(x)=G$.
So the only possible case is when $G$ is the trivial group.
Your approach is good, though, and is essentially a shorter version of my argument. The fact that the order of a conjugacy class is a divisor of the order of the group should be mentioned anyway.