I'm having some trouble with the following question:
Let $G$ be a group of order $p^n$ such that $|Z(G)| = p$ for some prime number $p$ and for $n\in \mathbb N_{\geq 2}$. Show that there exists an $a \in G$ such that $|C(a)| = p^{n-1}$.
I think that to solve this we use the fact that: $$|G|=|Z(G)| + \sum _{i=1}^k [G:C(x_i)]$$
for some $x_1,x_2,...,x_k$, but I don't see how to continue the proof from here. How can this be done?
The Class Equation yields: $$p^n=p+\sum_{i=1}^k\frac{p^n}{p^{\alpha_i}}$$ where $1<\alpha_i< n$, because for noncentral $x_i$: $Z(G)<C_G(x_i)<G$. Therefore: $$p^{n-1}=1+\sum_{i=1}^kp^{n-\alpha_i-1}$$ Since $p$ divides the LHS (as $n\ge 2$), if $n-\alpha_i-1>0$ for every $i$ we have a contradiction, because then $p\mid \sum_{i=1}^kp^{n-\alpha_i-1}$ but $p\nmid 1$. So, necessarily $n-\alpha_i-1=0$ for some $i$, namely $\alpha_i=n-1$ for some $i$.