p-groups and 2-generated abelian images Let $p$ be a prime number. Is there a finite nonabelian $p$-group $G$ such that any finite epimorphic $2$-generated image of $G$ is abelian?
 A: Let $G$ be any finite $p$-group, whose center and derived subgroup both have order $p$. Then every proper quotient of $G$ is abelian. In particular, if $G$ is not generated by 2 elements, it answers the question.
For instance, let $H$ be any non-abelian group of order $p^3$, denote by $Z$ its center. Define $G$ as the quotient of $H\times H$ by the diagonal of $Z\times Z$. Then $G$ has order $p^5$ satisfies the above conditions; its minimal number of generators is 4.
A: If a two-generator nonabelian p-group G satisfies the stated property, it is either of order p^3 or of class 2 with a cyclic subgroup of index p. To complete the classification, one can apply induction on |G|. I feel that we obtain the above mentioned groups.
A: Some time ago I asked to classify the nonabelian $p$-groups of exponent $>p$ containing exactly $p$ maximal abelian subgroups of exponent $>p$. Below I'l prove that if $G$ is such a group, then $\exp(G)=p^2$. Below we cite Berkovich-Janko, Groups of prime Power Order, 1,2.
Assume that $\exp(G)>p^2$. Then there is $x\in G$ of order $>p^2$. Let $B_1,\dots,B_p$ be all maximal abelian subgroups of $G$ of exponent $>p$. One may assume that $x\in B_1$ so that $\exp(B_1)>p^2$. As all maximal abelian subgroups cover $G$, there is in $G$ a maximal abelian subgroup $T$ of exponent $p$ (indeed, a nonabelian $p$-group is not covered by $p$ proper subgroups). Suppose that $T\triangleleft G$. Set $H=\langle x,T\rangle$. By Lemma 57.1, there is $t\in T$ such that $S=\langle t,x\rangle$ is minimal nonabelian. As $\exp(S)>p^2$ and $p>2$, there is in $S$ at least $p+1$ maximal abelian subgroups, say $E_1,\dots,E_{p+1}$, of exponent $>p$ (Lemma 65.1). Let $E_i\le U_i\le G$, where $U_i$ is maximal abelian in $G$. Then $U_1,\dots,U_{p+1}$ are pairwise distinct maximal abelian subgroups of $G$ of exponent $>p$, contrary to the hypothesis. Thus, all maximal abelian subgroups of $G$ of exponent $p$ are nonnormal. The same argument shows that the minimal nonabelian subgroup $S$ has at most $p$ maximal subgroups of exponent $>p$ hence $\exp(S)\le p^2$ (Lemma 65.1). Since $G$ has a normal maximal abelian subgroup (by the above, this is $B_i$ for some $i$), it follows that all $B_i\triangleleft G$. Set $F=B_1B_2$. Then $F\triangleleft G$ is of class $2$  (Fitting) and exponent $>p^2$. By Exercise 1.6, $F$ has exactly $p$ maximal abelian subgroups, say $L_1,\dots,L_p$, of exponent $>p$. Therefore, by Exercise 1.6 again, it has a maximal abelian subgroup $V$ of exponent $p$. The subgroup $V\triangleleft F$ since $\text{cl}(F)=2$. Therefore, by Lemma 57.1, there is $v\in V$ such that $K=\langle x,v\rangle$ is minimal nonabelian. As $\exp(K)>p^2$, we get a contradiction.
