Equivalence of definitions of gyrogroup

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This article gives a proof/explanation of the equivalence of multiple definitions for the term gyrogroup
View a complete list of pages giving proofs of equivalence of definitions

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:

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]

Related facts

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.