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Secret
Secret
12:44
(Revised) Corollary 3.2: Zero inverse are permutations in finite structures: Given an associative division by zero algebra $S$ with an underlying set of cardinality $n \in \mathbb{N}$. If there is a left (right) zero inverse $c$ and left (right) identity $1$, then the semigroup action of $c$ on $S$ is a permutation.
Secret
Secret
13:08
Proof: Let $0a=b$ be a left zero term. By Corollary 3.1, $0a$ is unique for some $a \in S$ (That is, the left semigroup action of $0$ on $a$ is injective).
Multiply both sides by $c$ to get $c0a=cb$. Using the left identity $a=cb$, which is also unique since $c$ is the left inverse of $0$. Hence the semigroup action of $c$ on $b$ is injective. It remains to prove that $c$ is surjective. Since $S$ is a semigroup, $(0a)1=0a \in S$ for all $a\in S$. Thus $0a$ exists for all $a \in S$. Now if $0$ is not surjective, then there exists $U \supseteq S$, that is the codomain of $0$. This is impossib…
 
3 hours later…
Secret
Secret
16:29
Theorem 8: Cyclic subgroup implies one sided null subsemigroup: Given an associative division by zero algebra $S$ containing a subgroup isomorphic to the cyclic subgroup $(\mathbb{Z}/n,+)$, then the elements in this subgroup induces a left or right null subsemigroup of the same size in the addition structure.
Secret
Secret
16:53
Proof: Since $X=(\mathbb{Z}/n,+)$ is a cyclic subgroup, it has unique inverses for all elements in the subgroup. Therefore, given any $a,b$ in the group such that $ab=c$, then $b=a^{-1}c$, thus $X$ satisfy the latin square property. Therefore, any left or right group action by element $a \in X$ induces a permutation to all elements. Now consider the left additive identity, $\forall s \in S$
$$0+s=s$$
Multiply both sides on the left by $k \in \{0,1,c\}$ and using left or right distributivity. Then
Corollary 3.3: Commutative zero terms implies cyclic subgroup of order $\geq 3$ Given an associative division by zero algebra $S$ of size $n$, where all zero terms commute, then it contains a cyclic subgroup of order at least $3$.
Secret
Secret
17:09
Proof: A division by zero algebra contains at least 3 elements, $0$, $1$ and the (possibly one sided) zero inverse $c$. Commutation of zero terms implies $\exists a,b \in S, a0=0a=b$. Suppose $0^2=c$, then by Theorem 7, $c0=0c=1$. By Corollary 3.2, the elements in $\{0,1,c\}$ are distinct permutations to this subset. Therefore this subset forms a cyclic subgroup of order 3.
Now suppose $0^2=d\neq c$, then by taking the left inverse $c$, $0=cd$. By Corollary 3.2, $\{0,1,c,d\}$ act on this subset as distinct permutations, thus they form a cyclic subgroup of order 4. The proof can be generalis…
Extra note: The fact that $\{0,1,c\}$ forms a cyclic subgroup prevent the structure from interacting with the rest of the algebra, resulting in them to be expressed as direct products of cyclic subgroups of order $3< x < n$ and some other structures $X$.
Theorem 8b: Decomposition theorem for finite associative division by zero algebras with commuting zero terms: Given a finite associative division by zero algebra $S$ with commuting zero terms, it is always possible to express them as a direct product of a semiring $R$ of the form where the multlplicative structure contains a cyclic subgroup of order $3 < x \leq n$ and some other ringnoids $X$. In symbols:$$S = R \times X$$
Theorem 6c: Division by zero no-go theorem (Commutativity) Associative division by zero with commuting zero terms are not interesting
Proof: Theorem 8b said the structure $S$ contains a cyclic subgroup of order at least 3, which by Theorem 8, must contain a one sided null semigroup of size at least 3. Therefore by the definition of not interesting, it is not interesting.
Theorem X1: Division by zero no-go theorem: Associative finte division by zero are not interesting

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