I' m trying to solve the following exercise:
Prove that, if G is finite of order $p^n$, p prime, $n\geq3$ and $|Z(G)|=p$, then G contains a conjugacy class of order $p$.
I know that every class of conjugacy must be of order $p^l$ where $l<n$ and that the sum of these orders plus $p=|Z(G)|$ must equal $p^n$, the class equation. We have the equation: $$p^n-p=p^{l_1}+p^{l_2}+\dots+p^{l_r}.$$
$r$ being the number of conjugacy classes in $G$. Since $p$ devides the LHS, it devides the RHS. I need to conclude that at least one of the $l_i=1$.
One more think that I know about the conjugacy classes, is that their order equals the index of the centralizer $C_G(x_i)$ in $G$, where $x_i$ is one representative of the conjugacy class ($x_i\notin Z(G)$.)
Since it is a subgroup of $G$, $|C_G(x_i)|=p^{k_i},$ $k_i<n$, $n-k_i=l_i$ and $(G:C_G(x_i))=p^{l_i}$.
So we could show that, the order of the centralizer of one of the representatives is equal to $p^{n-1}$.
I can't conlude, can somebody help me ?
Best Answer
Divide the LHS and RHS by $p$. Since $p$ does not divide the new LHS, it cannot divide the new RHS, thus your conclusion.