# 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 $M(p^n)$)

-
If $H$ is the p-group of exponent $p$ and order $p^3$, and $A$ is any abelian group, then any central product of $A$ and $H$ has center $A$ of index $p^2$. – Steve D Feb 21 '13 at 19:16
Steve, For any integer $n\geq 4$ there exist only one group sa you make. – Hamid Shahverdi Feb 21 '13 at 19:37
@Hamid: No, there's many such groups, obtained by varying $A$. E.g., with $n=5$, you can have $A=C_{p}^3$, yielding a group of exponent $p$, or $A=C_{p^3}$, yielding a group with an element of order $p^3$. – Arturo Magidin Feb 21 '13 at 19:59
Yes, You are right. – Hamid Shahverdi Feb 21 '13 at 20:30
A $p$-group $G$ satisfies the condition iff $G=S{\rm Z}(G)$, where $S\le G$ is minimal nonabelian. – yakov Jun 25 at 16:11

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.

-
Ralph, It means that there exist only one group of order $p^n$ for $n\geq 4$ with the property in the question? – Hamid Shahverdi Feb 21 '13 at 20:02
No it doens't. Potentially you get a different group for every choice of Z,a,b and c. – Johannes Hahn Feb 21 '13 at 20:31
And this is really as detailed as you can be. – Steve D Feb 21 '13 at 20:37
Thank you Ralph – Hamid Shahverdi Feb 22 '13 at 7:52

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:

1. $G/Z(G)\cong G/Z(K)$; and
2. $[G,G]\cong [K,K]$; and
3. 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 $p$.

-
Wait a minute... what map from $G/Z(G)\times G/Z(G)$ onto $[G,G]$ are you talking about? The commutator? That isn't onto in general. Also why should $[G,G]$ be cyclic of order $p$? Could you please eloborate on that? Thanks. – Johannes Hahn Feb 25 '13 at 14:48
@Johannes Hahn: The map has image which generates $[G,G]$; since the bilinear form is alternating, in the case at hand the image is generated by $[x,y]$ (where $G/Z(G)$ is generated by the images of$x$ and $y$), hence $[G,G]$ is cyclic; and since the map is bilinear, it will perforce be onto in this situation (rather than merely mapping onto a generating set). Since $G/Z(G)$ is of exponent $p$, $[x,y]$ is of order $p$, and so $[G,G]$ is cyclic of order $p$. – Arturo Magidin Feb 25 '13 at 15:30

See two my comments above containing solution of the problem. I repeat: A nonabelian $p$-group $G$ has the center of index $p^2$ if and only if $G=S\text{Z}(G)$, where $S\le G$ is minimal nonabelian.

-
You could have written the contents of your comments in your answer. Second, that you provide a short characterization does not necessarily mean that other answers are superfluous. In particular, a complete solution includes a list, like in Ralph's answer, and unlike in yours. – YCor Jul 5 at 12:40
@yakov Again, comments are for comments, not for answers. Comments are not permanent. They can be deleted at any time. The method of hide and seek is not an acceptable proof technique here. Put them in an answer. – zibadawa timmy Jul 20 at 20:32