Equivalence of definitions of LUCS-Baer Lie group

Statement
Suppose $$G$$ is a group of nilpotency class two (i.e., its derived subgroup is contained in ts center). Then, the following are equivalent:


 * 1) Every element of the derived subgroup $$G'$$ has a unique square root in $$G$$.
 * 2) Every element of its derived subgroup $$G'$$ has a unique square root among the elements in the center $$Z(G)$$.
 * 3) Every element of the derived subgroup $$G'$$ has a unique square root in $$G$$ and that square root is in the center $$Z(G)$$.

Facts used

 * 1) uses::Center is local powering-invariant
 * 2) uses::Equivalence of definitions of nilpotent group that is torsion-free for a set of primes

(3) implies (1)
This is immediate.

(3) implies (2)
If the square root is unique in all of $$G$$ and also happens to be in the center, it must be unique among the elements of the center as well.

(1) implies (3)
Given: A group $$G$$, an element $$g \in G'$$ such that there is a unique element $$x \in G$$ satisfying $$x^2 = g$$.

To prove: $$x \in Z(G)$$

Proof: Since $$G$$ has nilpotency class at most two, $$g \in G' \le Z(G)$$, so $$g \in Z(G)$$. The uniqueness of the square root, combined with Fact (1), tells us that $$x \in Z(G)$$ as well.

(2) implies (3)
Given: A group $$G$$ with the property that every element of $$G'$$ has a unique square root in $$Z(G)$$. An element $$g \in G'$$ with square root $$x \in Z(G)$$.

To prove: $$x$$ is the unique square root of $$g$$ in all of $$G$$.

Proof: Since every element of $$G'$$ has a unique square root in $$Z(G)$$, this in particular implies that the identity element has a unique square root in $$Z(G)$$, so $$Z(G)$$ is 2-torsion-free. By Fact (2) combined with $$G$$ being nilpotent, this implies that the squaring map is injective from $$G$$ to itself, so $$x$$ is the unique square root of $$g$$ in all of $$G$$.