nilpotent subgroup and hypercenter Let $G$ be a group and $Z^*(G)$ be it hypercenter. Suppose $z_1,z_2 \in Z^*(G)$ and $x \in G\setminus Z^*(G)$. Is it true that the group $\langle xz_1,xz_2 \rangle$ always generate a nilpotent subgroup? Thanks.
 A: Lemma: If $H \leq G$, then for $Z^n$ the members of the upper central series, we have $Z^n(G) \cap H \leq Z^n(H)$.
Proof: If $z \in Z(G) \cap H$ and $h \in H$, then $hz=zh$ since $h \in G$ and $z \in Z(G)$. Since $z$ commutes with $H$, $z \in Z(H)$. In other words, $Z(G) \cap H \leq Z(H)$. Suppose for induction that $Z^n(G) \cap H \leq Z^n(H)$, and that $z \in Z^{n+1}(G) \cap H$ and $h \in H$. Then $[h,z] \in Z^n(G)$ since $h \in G$ and $z \in Z^{n+1}(G)$. However, $[h,z] \in H$ since $h,z \in H$, so $[h,z]\in Z^n(G) \cap H \leq Z^n(H)$. Hence $z \in Z^{n+1}(H)$ because $[h,z] \in Z^n(H)$ for all $h \in H$. The induction proceeds. Taking unions, this also holds for limit ordinals, but we don't need that for finite groups. $\square$
Lemma: If $H/Z^n(H)$ is cyclic, then $H=Z^n(H)$ is nilpotent (assuming $n$ is finite).
Proof: Let $H/Z^n(H) = \langle g Z^n(H) \rangle$. Now consider any $h \in H$. Since $h Z^n(H) \in H/Z^n(H) = \langle g Z^n(H) \rangle$ we get $h=g^i z$ for some integer $i$ and element $z \in Z^n(H)$. Thus $[h,g] = [g^iz,g] = [g^i,g]^z [z,g] = 1^z [z,g] = [z,g] \in Z^{n-1}(H)$, where we use that $[ab,c]=[a,c]^b[b,c]$ and $[g^i,g]=1$. Hence $g \in Z^n(H)$ since $[h,g]\in Z^{n-1}(H)$ for all $h \in H$. Thus $gZ^n(H)=Z^n(H)$ and $H/Z^n(H)$ is trivial. In other words, $H=Z^n(H)$. $\square$
Proposition: If $x \in G$ and $z \in Z^*(G)$, then $\langle x,z \rangle$ is nilpotent.
Proof: Set $H=\langle x,z\rangle$. Then $z\in Z^n(G) \cap H \leq Z^n(H)$ and so $H/Z^n(H) = \langle x Z^n(H) \rangle$ is cyclic, and so $H=Z^n(H)$ is nilpotent. $\square$
Corollary: If $x \in G$ and $z_1, z_2 \in Z^*(G)$, then $\langle xz_1, xz_1 \rangle$ is nilpotent.
Proof: $\langle xz_1, xz_2 \rangle = \langle xz_1, z_1 z_2^{-1} \rangle$, but $xz_1 \in G$ and $z_1 z_2^{-1} \in Z^*(G)$. $\square$
