Let $G$ be a group of order $p^n$ where $p$ is prime. WITHOUT USING QUOTIENTS Prove that the center of $G$ cannot have order $p^{n-1}$. Let $G$ be a group of order $p^n$ where $p$ is prime.  Prove that the center  of $G$ cannot have order $p^{n-1}$ without using quotients.  The book I'm reading hasn't covered quotients or the meaning of the notation $A/B$ yet.
I tried by supposing the center is of order $p^{n-1}$, which implies there's $\phi(p^n)$ elements that aren't Abelian, but I'm not sure this is a good path.
 A: Let $G$ be a group with $Z(G)\ne G$ (in particular, $G$ is not abelian) and pick $a\in G\setminus Z(G)$.
Let $$A=\{\,a^kz\mid k\in\Bbb Z, z\in Z(G)\,\}.$$
Then $A$ is an abelian (hence proper) subgroup of $G$ because $a^kz\cdot a^{k'}z'=a^{k+k'}zz'$ and of course also a proper supergroup of $Z(G)$.
By multiplicativity of index, $$[G:Z(G)] =\underbrace{[G:A]}_{>1}\cdot \underbrace{[A:Z(G)]}_{>1}$$
cannot be prime.
If you do not have multiplacivity of index yet (or maybe not even index), you may just as well simply use that $|A|$ is a proper multiple of $|Z(G)|$ and a proper divisor of $|G|$. Or that $|A|=m\cdot |Z(G)|$ for some $m$ dividing the order of $a$, which again must divide $|G|$ - so that $m\ge p$.
A: As Cave Johnson has said, you can just rewrite the proof pretending you don't know about quotients.
Let $Z(G)$ be of order $p^{n-1}$.
Take an element $a\not\in Z(G)$ and consider it's order. We know that $a$ has to have prime order by Lagrange. If $\{z_1,...,z_{p^{n-1}}\}=Z(G)$(distinct elements) we claim that $A_i=\{a^i z_1,...,a^i z_{p^{n-1}}\}$ for $i=1,...,p$ are pairwise disjoint and  are all of order $p^{n-1}$. 
(Have you seen cosets? I presume you must have because you need it in the proof of Lagrange. In that case, the cosets $aZ(G),...,a^p Z(G)=Z(G)$ are pairwise disjoint and also all of size $p^{n-1}$. If you haven't the reason why they are pairwise disjoint is because if two elements $a^i z_m=a^j z_n$ then $a^{i-j}=z_n z_m^{-1})$ and if $i$ nad $j$ are distinct we can apply Bezout's theorem from number theory to invert $i-j$ to get an expression for $a$ in terms of elements in $Z(G)$)
We therefore have listed all the elements of $G$. Take two elements of $G$ and we see htat
$a^i z_m a^j z_n=a^i a^jz_m z_n =a^j a^i z_n z_m=a^j z_n a^i z_m$
(Try and justify to yourself why each of these parts commuted.)
