# Cardinality of Center of p Groups $|Z(G)| \neq p^{n-1}$

Question

Let G be a p group, such that $|G|=p^n$. Prove that $|Z(G)|\neq p^{n-1}$.

Beginning of Proof

So, I began by writing out $p^n=|G|=\sum_{i\in I}[G:Stab_G(x_i)]+|Z(G)|=\sum_{i\in I}[G:Stab_G(x_i)]+p^{n-1}$, when $\left\{x_i\right\}_{i\in I}$ are representatives in the following way $\cup_{i\in I}x_iGx_i^{-1}\cup Z(G)=\cup_{x\in G}xGx^{-1}$, when $\forall i\in I, |x_iGx_i^{-1}|\geq p$ (because otherwise $x_i \in Z(G)$). Therefore, $\sum_{i\in I}[G:Stab_G(x_i)]=p^n(p-1)$.

Then, we can say that $\forall i\in I, \exists j_i\in [1,n-1]: [G:Stab_G(x_i)]=p^{j_i}$, according to Lagrange's theorem. Since p-1 is even (if $p\neq 2$), then in order for this sum of odd numbers to be even, then |I| must be even. Therefore, we can write that $|I|=2k$, when $2k \in [2,\frac{p^n(p-1)}{2}]$. $2k\neq p^n(p-1)$, because then, we'll have $p^n(p-1)$ numbers in the sum, and therefore every number in the sum must be 1, such that $\forall i\in I, |Stab_G(x_i)|=|G| \Rightarrow x_i\in Z(G)$. Therefore, $k\in [1,\frac{p^n(p-1)}{4}]$, and therefore, $p-1\equiv 0 (mod 4) \Rightarrow p \equiv 1 (mod 4)$. Therefore, $|Z(G)|\neq p^{n-1}$, for $p: p \equiv 3(mod 4)$.

I haven't been able to prove this for all other primes p. Thanks a lot for the help!!!

Hint: If $|G| = p^n$ and $|Z(G)| = p^{n-1}$, $|G/Z(G)| = p$. Now take a look at If $G/Z(G)$ is cyclic, then $G$ is abelian. What can you conclude now?
• :CONCLUSION:From this we get,$Z(G)=G \implies \vert Z(G) \vert = \vert G \vert$ which is a contradiction to the hypothesis $\vert Z(G) \vert=p^{n-1}$. Is it so?. – Styles Jul 30 '16 at 8:06
You need know some basic facts. If $G$ is abelian so $G=Z(G)$. But $|G|\neq|Z(G)|$. So assume that $G$ is non-abelian. Since $Z(G)$ is a normal subgroup so $G/Z(G)$ is defined and $[G:Z(G)]=p$. In this case using another answer hint you get $G$ is abelian which is a contradiction.
• @amWhy: I am glad cause of that. This is more precious to me that I can make someone around the World to have a good feeling. This shows that $0$ (it is me) is not nothing. :-) – Mikasa May 23 '13 at 19:05