ON CAPABLE GROUPS OF ORDER p4

Abstract

A group $H$ is said to be capable, if there exists another group
$G$ such that $\frac{G}{Z(G)}~\cong~H$, where $Z(G)$ denotes the
center of $G$. In a recent paper \cite{2}, the authors
considered the problem of capability of five   non-abelian $p-$groups of order $p^4$ into account. In this paper, we continue this paper by considering three other groups of order $p^4$.  It is proved that the group $$H_6=\langle x, y, z \mid x^{p^2}=y^p=z^p= 1, yx=x^{p+1}y, zx=xyz, yz=zy\rangle$$ is not capable. Moreover, if $p > 3$ is  prime and $d \not\equiv 0, 1 \ (mod \ p)$ then the following groups are not capable:\\
{\tiny $H_7^1=\langle x, y, z \mid x^{9} = y^3 = 1, z^3 = x^{3}, yx = x^{4}y, zx = xyz, zy = yz \rangle$,\\
$H_7^2= \langle x, y, z \mid x^{p^2} = y^p = z^p = 1, yx = x^{p+1}y, zx = x^{p+1}yz, zy = x^pyz \rangle,$ \\
$H_8^1=\langle x, y, z \mid x^{9} = y^3 = 1, z^3 = x^{-3}, yx = x^{4}y, zx = xyz, zy = yz \rangle$,\\
$H_8^2=\langle x, y, z \mid x^{p^2} = y^p = z^p = 1, yx = x^{p+1}y, zx = x^{dp+1}yz, zy = x^{dp}yz \rangle$.}