# Equivalence of definitions of gyrogroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term gyrogroup
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## The definitions that we have to prove are equivalent

We want to prove that the following two definitions are equivalent.

### Minimal definition

A magma with underlying set $G$ and binary operation $*$ is termed a gyrogroup if the following hold:

• Left identity and left inverse: There is an element $e \in G$ such that $e$ is a left neutral element and every element has a left inverse with respect to $e$. In other words:

$e * a = a \forall a \in G$

and for all $a \in G$, there exists $b \in G$ such that:

$b * a = e$

• Gyroassociativity: For any $a,b,c \in G$, there is a unique element $\operatorname{gyr}[a,b]c \in G$ such that:

$a * (b * c) = (a * b) * (\operatorname{gyr}[a,b]c)$

• Gyroautomorphism: $\operatorname{gyr}[a,b]$ (i.e., the map that sends $c$ to $\operatorname{gyr}[a,b]c$) is a magma automorphism of $G$. This is called the Thomas gyration, or gyroautomorphism, of $G$.
• Left loop property: The following are equal as automorphisms of $G$:

$\operatorname{gyr}[a,b] = \operatorname{gyr}[a * b,b]$

### Maximal definition

A magma with underlying set $G$ and binary operation $*$ is termed a gyrogroup if the following hold:

• Two-sided identity and two-sided inverse: There is a unique element $e \in G$ such that $e$ is a two-sided neutral element and every element has a unique two-sided inverse element with respect to $e$. In other words:

$e * a = a * e = a\forall a \in G$

and for all $a \in G$, there exists a unique two-sided inverse $b \in G$ such that:

$b * a = a * b = e$

The element $b$ is denoted $a^{-1}$.

• Gyroassociativity: For any $a,b,c \in G$, there is a unique element $\operatorname{gyr}[a,b]c \in G$ such that:

$a * (b * c) = (a * b) * (\operatorname{gyr}[a,b]c)$

• Gyroautomorphism: $\operatorname{gyr}[a,b]$ (i.e., the map that sends $c$ to $\operatorname{gyr}[a,b]c$) is a magma automorphism of $G$. This is called the Thomas gyration, or gyroautomorphism, of $G$.
• Left loop property: The following are equal as automorphisms of $G$:

$\operatorname{gyr}[a,b] = \operatorname{gyr}[a * b,b]$

## Facts used

1. Binary operation on magma determines neutral element, which follows from equality of left and right neutral element

## Proof

We will show that the minimal definition implies the maximal definition, i.e., the existence of a left neutral element and left inverses, along with the other axioms, forces the left neutral element to be the unique two-sided neutral element and forces the left inverse to be the unique two-sided inverse.

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### Proof of two-sided neutral element and two-sided inverse

Given: A gyrogroup $(G,*)$ (as per the minimal definition) with left neutral element $e$.

To prove: $e$ is a two-sided neutral element and every element has a two-sided inverse. Explicitly, for all $a \in G$, we have $e * a = a * e = a$, and there exists $b \in G$ such that $b * a = a * b = e$.

Proof: We fix $a \in G$.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $\operatorname{gyr}[e,x]$ is the identity map for all $x \in G$. We note that, for any $y \in G$, $e * (x * y) = x * y = (e * x) * y$. By uniqueness of $\operatorname{gyr}[e,x]y$, we get $\operatorname{gyr}[e,x]y = y$.
2 Suppose $x,z \in G$ are such that $z * x = e$. Then, $\operatorname{gyr}[z,x]$ is the identity map. gyroassociativity Step (1) By the left loop property, $\operatorname{gyr}[z,x] = \operatorname{gyr}[z * x,x] = \operatorname{gyr}[e,x]$ which is the identity map by Step (1).
3 Recall the element $a$ fixed beforehand. Let $b$ be a left inverse of $a$ and $c$ be a left inverse of $b$. In other words, $c * b = b * a = e$. existence of left inverses
4 $\operatorname{gyr}[b,a] = \operatorname{gyr}[c,b]$ and both equal the identity map of $G$. Steps (2), (3) Step-combination direct.
5 For all $x \in G$, $c * (b * x) = x$ and $b * (a * x) = x$. Steps (3), (4) By Step (4), $c * (b * x) = (c * b) * x$. By Step (3), this becomes $e * x = x$. Similarly for $b * (a * x)$.
6 $c * (b * (a * e)) = a * e$. Step (5) Use $x = a * e$ in Step (5) for the $c * (b * x) = x$ part.
7 $c * (b * (a * e)) = c * e$. Step (5) Use $x = e$ in Step (5) for the $b * (a * x) = x$ part.
8 $c * e = c * (b * a) = a$. Step (5) Rewrite $e = b * a$ to get $c * e = c * (b * a)$. Then, use Step (5) setting $x = a$ in $c * (b * x) = x$.
9 $a * e = a$. Since $e$ is already known to be left neutral, we get $e * a = a * e = a$. Steps (6), (7), (8) Step-combination direct.
10 Note that $a$ was fixed but arbitrary for the proof up to Step (9). Thus, the same proof could be given for $c$ in place of $a$ and we would get $c * e = c$. Step (9) direct
11 $a = c$ Steps (8), (10) Step-combination direct
12 $b * a = a * b = e$. Steps (3), (11) Step-combination direct
13 Steps (9) and (12) complete the proof.

### Proof of uniqueness of neutral element

This is direct from Fact (1) -- a two-sided neutral element, if it exists, must be unique.

### Proof of uniqueness of two-sided inverse

We will show something stronger, any left inverse must equal any right inverse.

Given: A gyrogroup $(G,*)$ with two-sided neutral element $e$. An element $a \in G$ with left inverse $b$ and right inverse $d$. In other words, $b * a = a * d = e$.

To prove: $b = d$

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $\operatorname{gyr}[e,x]$ is the identity map for all $x \in G$. We note that, for any $y \in G$, $e * (x * y) = x * y = (e * x) * y$. By uniqueness of $\operatorname{gyr}[e,x]y$, we get $\operatorname{gyr}[e,x]y = y$.
2 Suppose $x,z \in G$ are such that $z * x = e$. Then, $\operatorname{gyr}[z,x]$ is the identity map. gyroassociativity Step (1) By the left loop property, $\operatorname{gyr}[z,x] = \operatorname{gyr}[z * x,x] = \operatorname{gyr}[e,x]$ which is the identity map by Step (1).
3 $\operatorname{gyr}[b,a]$ is the identity map. Thus, $b * (a * d) = (b * a) * d$. $b * a = e$. Step (2)
4 We get $b = d$ $b * a = a * d = e$ Step (3) The result follows directly from simplifying Step (3), using $b * a = a * d = e$ is a two-sided neutral element.