Gyrogroup implies left-inverse property loop

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Suppose (G,*) is a gyrogroup with neutral element e and inverse map {}^{-1}. Then, (G,*) is a left inverse property loop with the same neutral element e and where the left inverse map is the same as the inverse map.


Given: A gyrogroup (G,*), elements a,b \in G.

To prove: There exists unique u \in G such that a * u = b and there exists unique y \in G such that v * a = b (together, these prove it's a loop). Further, u = a^{-1} * b (this shows the left inverse property).


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}[a^{-1},a] is equal to the identity map. Step (2)
4 We have a * u = b \implies a^{-1} * (a * u) = a^{-1} * b. This simplifies to u = a^{-1} * b. Step (3) By Step (3), we obtain that a^{-1} * (a * u) = (a^{-1} * a) * u and thus it simplifies to e * u = u.
5 v * a = b \implies \operatorname{gyr}[v,a] = \operatorname{gyr}[b,a] left loop property we use that \operatorname{gyr}[v,a] = \operatorname{gyr}[v * a,a]
6 We have v * a = b \implies v = b * \operatorname{gyr}[b,a](a^{-1}). In particular, v exists and is uniquely determined by a,b. Step (5) Consider the gyroassociativity of v,a,a^{-1}: v * (a * a^{-1}) = (v * a) * \operatorname{gyr}[v,a]a^{-1}. Simplifying the left side gives v = (v * a) * \operatorname{gyr}[v,a]a^{-1}. Now use Step (5) and we are done.

Steps (4) and (6) complete the proof.