Equivalence of definitions of gyrogroup
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
Contents
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 and binary operation is termed a gyrogroup if the following hold:
- Left identity and left inverse: There is an element such that is a left neutral element and every element has a left inverse with respect to . In other words:
and for all , there exists such that:
- Gyroassociativity: For any , there is a unique element such that:
- Gyroautomorphism: (i.e., the map that sends to ) is a magma automorphism of . This is called the Thomas gyration, or gyroautomorphism, of .
- Left loop property: The following are equal as automorphisms of :
Maximal definition
A magma with underlying set and binary operation is termed a gyrogroup if the following hold:
- Two-sided identity and two-sided inverse: There is a unique element such that is a two-sided neutral element and every element has a unique two-sided inverse element with respect to . In other words:
and for all , there exists a unique two-sided inverse such that:
The element is denoted .
- Gyroassociativity: For any , there is a unique element such that:
- Gyroautomorphism: (i.e., the map that sends to ) is a magma automorphism of . This is called the Thomas gyration, or gyroautomorphism, of .
- Left loop property: The following are equal as automorphisms of :
Related facts
- Semigroup with left neutral element where every element is left-invertible equals group
- Monoid where every element is left-invertible equals group
Facts used
- 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 (as per the minimal definition) with left neutral element .
To prove: is a two-sided neutral element and every element has a two-sided inverse. Explicitly, for all , we have , and there exists such that .
Proof: We fix .
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | is the identity map for all . | We note that, for any , . By uniqueness of , we get . | |||
2 | Suppose are such that . Then, is the identity map. | gyroassociativity | Step (1) | By the left loop property, which is the identity map by Step (1). | |
3 | Recall the element fixed beforehand. Let be a left inverse of and be a left inverse of . In other words, . | existence of left inverses | |||
4 | and both equal the identity map of . | Steps (2), (3) | Step-combination direct. | ||
5 | For all , and . | Steps (3), (4) | By Step (4), . By Step (3), this becomes . Similarly for . | ||
6 | . | Step (5) | Use in Step (5) for the part. | ||
7 | . | Step (5) | Use in Step (5) for the part. | ||
8 | . | Step (5) | Rewrite to get . Then, use Step (5) setting in . | ||
9 | . Since is already known to be left neutral, we get . | Steps (6), (7), (8) | Step-combination direct. | ||
10 | Note that was fixed but arbitrary for the proof up to Step (9). Thus, the same proof could be given for in place of and we would get . | Step (9) | direct | ||
11 | Steps (8), (10) | Step-combination direct | |||
12 | . | 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 with two-sided neutral element . An element with left inverse and right inverse . In other words, .
To prove:
Proof:
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | is the identity map for all . | We note that, for any , . By uniqueness of , we get . | |||
2 | Suppose are such that . Then, is the identity map. | gyroassociativity | Step (1) | By the left loop property, which is the identity map by Step (1). | |
3 | is the identity map. Thus, . | . | Step (2) | ||
4 | We get | Step (3) | The result follows directly from simplifying Step (3), using is a two-sided neutral element. |