# Equivalence of definitions of LUCS-Baer Lie group

This article gives a proof/explanation of the equivalence of multiple definitions for the term LUCS-Baer Lie group
View a complete list of pages giving proofs of equivalence of definitions

## 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 $Z(G)$.

## Proof

### (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 $G$.