p-group with large center Is there any characterization for $p$-groups of order greater than $p^3$ which center has index $p^2$? (One group whit this property if $M(p^n)$)
 A: This is much less informative than Ralph's excellent answer, but a quick observation is that, for $p\gt 2$, the groups you are looking for are exactly the groups that are isoclinic to the nonabelian groups of order $p^3$. This observation was made by Philip Hall in The classification of prime-power groups, J. Reine Angew. Math. 182 (1940) 130-141; the observation can be found at the bottom of 136.
Recall that two groups $G$ an $K$ are isoclinic if and only if:


*

*$G/Z(G)\cong G/Z(K)$; and

*$[G,G]\cong [K,K]$; and

*The isomorphisms can be chosen to be compatible; that is, if $\alpha\colon G/Z(G)\cong K/Z(K)$ and $\beta\colon [G,G]\cong[K,K]$, then for all $g,g'\in G$, $$\beta([gZ(G),g'Z(G)]) = [\alpha(gZ(G)),\alpha(g'Z(G))].$$


The central quotient of your groups are of order $p^2$, and since a nontrivial cyclic group cannot be isomorphic to a central quotient, the central quotient is isomorphic to $C_p\times C_p$; the commutator subgroup is contained in the center, and so the group is of class $2$. Since there is a bilinear alternating map from $(G/Z(G))\times (G/Z(G))$ onto $[G,G]$. Added: in general, the map $(G/Z(G))\times (G/Z(G))\to[G,G]$ given by $(xZ(G),yZ(G))\mapsto [x,y]$ has image that generates $[G,G]$; in this case, since $G$ is of class $2$, the map is bilinear; if $G/Z(G)$ is generated by $x$ and $y$, then it follows that $[G,G]$ is generated by $[x,x]$, $[x,y]$, $[y,x]$, and $[y,y]$; the first and last are trivial, and third equals the inverse of the second, so $[G,G]$ is cyclic generated by $[x,y]$. It now follows that the map is in fact onto. Since $G/Z(G)$ is of exponent $p$, then so is $[G,G]$ by the bilinearity of the bracket. Since $G$ is nonabelian, it follows that $[G,G]$ is cyclic of order $p$. It is now straightforward to see that a group with center of index $p^2$ is necessarily isoclinic to the nonabelian groups of order $p^3$. Conversely, if $G$ is isoclinic to the nonabelian groups of order $p^3$, then their central quotients must be of order $p^2$, giving the equivalence.
An alternative description in the case of $p$-groups, also given by Hall, is that they are precisely the nonabelian $p$-groups that have at least two abelian subgroups of index $p$. 
A: Let $G$ be a group in question. First note that $G/Z(G)\cong C_{p^2}$ isn't possible. Thus $G$ has the presentation 
$$\langle Z,x,y\mid Z \text{ central}, x^p=a,y^p=b,[x,y]=c\rangle$$
where $Z$ is the center, $a,b, c \in Z$ and $c\neq 1,\; c^p=1$. 
Added: Suppose $a=a_0^p,a_0 \in Z$. By replacing $x$ by $xa_0^{-1}$ we have the relation $x^p = 1$. Write $Z=\langle z_1,...,z_n\mid r(Z)\rangle$. Then we obtain the presentations 
$$\tag{I}G(c)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=y^p=1,[x,y]=c\rangle$$
$$G(c,i)= \langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i, y^p=1,[x,y]=c\rangle\tag{II}$$
$$G(c,i,j)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i,y^p=z_j,[x,y]=c\rangle\tag{III}$$
Added 2: 1) In case (III) we can assume $i\neq j$ (otherwise,  replacing $x$ by $xy^{-1}$ gives case (II)). 
2) Let $\exp(z_i)=k_i$. Denote by $(k_1,...,k_n)$ the isomorphism type of $C_{p^{k_1}} \times \cdots C_{p^{k_n}}$. Then the groups above have maximal abelian subgroups of the following types: 
$$\begin{array}{lcl}
(I)   & : & (k_1,...,k_n,1) \newline 
(II)  & : & (k_1,...,k_n,1), (k_1,..,k_i+1,..,k_n) \newline 
(III) & : & (k_1,..,k_i+1,..,k_n), (k_1,..,k_j+1,..,k_n)
\end{array}$$
Hence (I), (II), (III) belong to different ismorphism types. 
It remains to check for which parameters $c,i,j$ the groups within (I) resp. (II) resp. (III) are isomorphic. 
Added 3: Modulo possible mistakes a complete classification is given by: 

Let $p$ be an odd prime, $Z=\langle z_1,...,z_n\rangle \cong C_{p^{l_1}}^{n_1} \times \cdots \times  C_{p^{l_m}}^{n_m}$ with $l_1 > \cdots > l_m$ and suppose 
$$\lbrace 1,...,n\rbrace = \coprod_{i=1}^m \lbrace r_i,...,r_{i+1}-1\rbrace\qquad (n_i=r_{i+1}-r_i)$$ 
is a decomposition such that $\exp(z_j)=l_i$ for $r_i \le j < r_{i+1}$. Then the groups $G$ with $Z(G)=Z$ and $(G:Z(G))=p^2$ are given (up to isomorphism) by the following non-isomorphic presentations $(c_i := z_{r_i}^{p^{l_i-1}}):$
$$\begin{array}{llcl}
G(c_i) & (i=1,...,m) & \qquad &  \text{(I)} \newline 
G(c_i,r_j) & (i,j=1,...,m) & \qquad &  \text{(II)} \newline 
G(c_i,r_j,r_j+1) &  (i,j=1,...,m,\;n_j \ge 2) & \qquad &  \text{(IIIa)} \newline 
G(c_i,r_j,r_k) & (i,j,k=1,...,m,\;j \neq k) & \qquad &  \text{(IIIb)}
\end{array}$$
In particular, there are $m^3 +m(1 + |\lbrace n_j \ge 2\rbrace|)$ isomorphism classes.  

Proof: i) It follows from the structure of the max. abelian subgroups in Add-2, that $G(c,i) \cong G(c,j)$ iff $k_i=k_j$ and $G(c,i,j)\cong G(c,l,q),\;(i\le j,\;l \le q)$ iff $k_i = k_l$ and $k_j=k_q$. 
ii) We determine the isomorphism types for various $c$.  
Claim 1: $\quad G(c,-) \cong G(c',-)$ iff there is $f \in Aut(Z)$ such that $c'=f(c)$. 
First, $f \in Aut(Z)$ extends to an isomorphism $\varphi_f: G(c,-) \to G(f(c),-)$ by $x \mapsto x, y\mapsto y$. Conversely, if $\varphi: G(c,-) \to G(c',-)$ is an isomorphism, then (easy calculation) there is $q$ coprime to $p$ with $c'=\varphi(c)^q$. Hence $c'=f(c)$ where $f\in Aut(Z)$ is the composition of $\varphi|Z$  and the $q$-power map. $\square$ 
Using $Z=\prod_{j=1}^m C_{p^{l_j}}^{n_j}$, we have the decomposition 
$\langle c\in Z \mid c^p=1\rangle = \coprod_{i=1}^m M_i$, where 
$$M_i = \lbrace (g_1,...,g_i,1,...,1)\mid g_i \neq 1,\;g_j \in C_{p^{l_j}}^{n_j},\; g_j^p=1\;(j=1,...,i)\;\rangle.$$ 
Claim 2: $\quad f(M_i)=M_i$ for $f \in Aut(Z)$. Conversely, for $c, c' \in M_i$ there is $f \in Aut(Z)$ such that $f(c)=c'$. 
I omit the proof which is in essential based on linear algebra. Since $c_i = z_{r_i}^{p^{l_i-1}} \in M_i$, the claim implies $G(c,-)\cong G(c_i,-)$ for all $c \in M_i$ and $G(c_i,-) \not\cong G(c_j,-)$ for $i\neq j$. This completes the classification.
Remark:  For $p=2$ (in contrast to $p$ odd) we have in addition the case $G(c,i,i)$ (i.e. $x^2=y^2=c\neq 1$). Example: Quaternion group of order 8. The reason is that $(xy^{-1})^p=c^{p(p+1)/2}$ is $1$ only if $p$ is odd.  
A: As this post contains a correct result, is it unclear why it is estimated by -4. This can do only unqualified person. Since only my posts estimated negatively, I decided to delete all them.
