Repost: Let p be a prime, let n>2 be an integer, and let G be a nonabelian group of order p^n. Prove the center of G cannot have order p^(n−1) Here is the original question:
Prove the center of $G$ cannot have order $p^{n-1}$
My question is regarding some of the hints, the sketch seems to suggest that the tactic is to show that |G|/|Z(G)|=p, then G\Z(G) is order p and therefore cyclic which is a contradiction as G is clearly non abelian. My questions are:
1) is G/Z(G) the group of cosets, similar to Z/nZ as representatives of infinite groups? 
2) Why does having order p insure that G/Z(G) is cyclic?
Thanks!
 A: 1) Yes, whenever $N$ is a normal subgroup of a group $G$, we can speak of the quotient group $G/N$, which is the set of cosets $aN$ with the operations $aN\cdot bN=abN$ (which works precisely because $N$ is normal).
2) Any non-neutral element of a group of prime order $p$ must have order $p$ (because the order must be a divisor of $p$ and $>1$) and hence generate the group. 
A: The answer to 1) is yes.  For any subgroup $H$, $G/H=\{xH:x\in G\}$, the set of left cosets $H$ in $G$.
For 2) since the order of a subgroup must divide the order of the original group, the only possible subgroups are $\{e\}$, where $e$ is the identity element, or $G$ itself.  So, what does that mean about a subgroup generated by a single element?
A: Suppose that the order of $Z(G)$ is $p^{n-1}$, $G/Z(G)$ is a cyclic group of order $p$.Let $q:G\rightarrow G/Z(G)$ the projection, and $x\in G$, such that $q(x)$ generates $G/Z(G)$, $y\in G$, $q(y)=q(x)^n$ implies that $y=ux^n$, $u\in Z(G)$. Let $z\in G$.
Write $z=vx^m, v\in Z(G)$, $[z,y]=[vx^m,ux^n]=1$ since $u,v\in Z(G)$ this is in contradiction with the fact that $G$ is non abelian.
