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What's the maximum probability of associativity for triples in a nonassociative loop?
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In a finite nonabelian group, the probability that two randomly chosen elements commute cannot exceed 5/8. One easy proof also makes it easy to find the smallest groups that attain this bound, namely the two nonabelian groups of order 8:
The 5/8 theorem.
This makes me wonder: how large can the probability be that three randomly chosen elements $a,b,c$ of a finite nonassociative loop associate, i.e. obey $(ab)c = a(bc)$?
You can prove the 5/8 theorem for groups by separately settling two questions:
What is the largest possible fraction of elements of a finite group that lie in the center? (Answer: 1/2)
Given a noncentral element of a finite group, what's the largest possible fraction of elements that commute with it? (Answer: 1/4)
The nonabelian groups of order 8 achieve both these upper bounds. We could try a similar strategy for my question, attempting to settle these:
What is the largest possible fraction of elements $a$ in a finite loop such that $(ab)c = a(bc)$ for all elements $b,c$?
If an element $a$ of a finite loop does not have $(ab)c = a(bc)$ for all elements $b,c$, what is the largest possible fraction of elements $b$ such that $(ab)c = a(bc)$ for all $c$?
If a pair $a,b$ does not have $(ab)c = a(bc)$ for all elements $c$, what is the largest possible fraction of elements $c$ such that $(ab)c = a(bc)$?
Unfortunately I don't know how to settle these.
Since the quaternion 8-group $$Q_8 = pm 1, pm i, pm j, pm k$$ attains the 5/8 bound for commutativity of pairs in a nonabelian group, one might hope that the octonion 16-loop $$O_16 = pm 1, pm e_1, dots, pm e_7$$ attains the maximum probability of associativity for triples in a nonassociative loop. Does it?
I'm afraid I haven't even worked out the probability that a triple in $O_16$ associates, though it would be easy to do.
finite-groups non-associative-algebras magmas
asked 3 hours ago
John Baez
8,0384189
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up vote
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In a finite nonabelian group, the probability that two randomly chosen elements commute cannot exceed 5/8. One easy proof also makes it easy to find the smallest groups that attain this bound, namely the two nonabelian groups of order 8:
The 5/8 theorem.
This makes me wonder: how large can the probability be that three randomly chosen elements $a,b,c$ of a finite nonassociative loop associate, i.e. obey $(ab)c = a(bc)$?
You can prove the 5/8 theorem for groups by separately settling two questions:
What is the largest possible fraction of elements of a finite group that lie in the center? (Answer: 1/2)
Given a noncentral element of a finite group, what's the largest possible fraction of elements that commute with it? (Answer: 1/4)
The nonabelian groups of order 8 achieve both these upper bounds. We could try a similar strategy for my question, attempting to settle these:
What is the largest possible fraction of elements $a$ in a finite loop such that $(ab)c = a(bc)$ for all elements $b,c$?
If an element $a$ of a finite loop does not have $(ab)c = a(bc)$ for all elements $b,c$, what is the largest possible fraction of elements $b$ such that $(ab)c = a(bc)$ for all $c$?
If a pair $a,b$ does not have $(ab)c = a(bc)$ for all elements $c$, what is the largest possible fraction of elements $c$ such that $(ab)c = a(bc)$?
Unfortunately I don't know how to settle these.
Since the quaternion 8-group $$Q_8 = pm 1, pm i, pm j, pm k$$ attains the 5/8 bound for commutativity of pairs in a nonabelian group, one might hope that the octonion 16-loop $$O_16 = pm 1, pm e_1, dots, pm e_7$$ attains the maximum probability of associativity for triples in a nonassociative loop. Does it?
I'm afraid I haven't even worked out the probability that a triple in $O_16$ associates, though it would be easy to do.
finite-groups non-associative-algebras magmas
asked 3 hours ago
John Baez
8,0384189
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In a finite nonabelian group, the probability that two randomly chosen elements commute cannot exceed 5/8. One easy proof also makes it easy to find the smallest groups that attain this bound, namely the two nonabelian groups of order 8:
The 5/8 theorem.
This makes me wonder: how large can the probability be that three randomly chosen elements $a,b,c$ of a finite nonassociative loop associate, i.e. obey $(ab)c = a(bc)$?
You can prove the 5/8 theorem for groups by separately settling two questions:
What is the largest possible fraction of elements of a finite group that lie in the center? (Answer: 1/2)
Given a noncentral element of a finite group, what's the largest possible fraction of elements that commute with it? (Answer: 1/4)
The nonabelian groups of order 8 achieve both these upper bounds. We could try a similar strategy for my question, attempting to settle these:
What is the largest possible fraction of elements $a$ in a finite loop such that $(ab)c = a(bc)$ for all elements $b,c$?
If an element $a$ of a finite loop does not have $(ab)c = a(bc)$ for all elements $b,c$, what is the largest possible fraction of elements $b$ such that $(ab)c = a(bc)$ for all $c$?
If a pair $a,b$ does not have $(ab)c = a(bc)$ for all elements $c$, what is the largest possible fraction of elements $c$ such that $(ab)c = a(bc)$?
Unfortunately I don't know how to settle these.
Since the quaternion 8-group $$Q_8 = pm 1, pm i, pm j, pm k$$ attains the 5/8 bound for commutativity of pairs in a nonabelian group, one might hope that the octonion 16-loop $$O_16 = pm 1, pm e_1, dots, pm e_7$$ attains the maximum probability of associativity for triples in a nonassociative loop. Does it?
I'm afraid I haven't even worked out the probability that a triple in $O_16$ associates, though it would be easy to do.
finite-groups non-associative-algebras magmas
asked 3 hours ago
John Baez
8,0384189
In a finite nonabelian group, the probability that two randomly chosen elements commute cannot exceed 5/8. One easy proof also makes it easy to find the smallest groups that attain this bound, namely the two nonabelian groups of order 8:
The 5/8 theorem.
This makes me wonder: how large can the probability be that three randomly chosen elements $a,b,c$ of a finite nonassociative loop associate, i.e. obey $(ab)c = a(bc)$?
You can prove the 5/8 theorem for groups by separately settling two questions:
What is the largest possible fraction of elements of a finite group that lie in the center? (Answer: 1/2)
Given a noncentral element of a finite group, what's the largest possible fraction of elements that commute with it? (Answer: 1/4)
The nonabelian groups of order 8 achieve both these upper bounds. We could try a similar strategy for my question, attempting to settle these:
What is the largest possible fraction of elements $a$ in a finite loop such that $(ab)c = a(bc)$ for all elements $b,c$?
If an element $a$ of a finite loop does not have $(ab)c = a(bc)$ for all elements $b,c$, what is the largest possible fraction of elements $b$ such that $(ab)c = a(bc)$ for all $c$?
If a pair $a,b$ does not have $(ab)c = a(bc)$ for all elements $c$, what is the largest possible fraction of elements $c$ such that $(ab)c = a(bc)$?
Unfortunately I don't know how to settle these.
Since the quaternion 8-group $$Q_8 = pm 1, pm i, pm j, pm k$$ attains the 5/8 bound for commutativity of pairs in a nonabelian group, one might hope that the octonion 16-loop $$O_16 = pm 1, pm e_1, dots, pm e_7$$ attains the maximum probability of associativity for triples in a nonassociative loop. Does it?
I'm afraid I haven't even worked out the probability that a triple in $O_16$ associates, though it would be easy to do.
finite-groups non-associative-algebras magmas
finite-groups non-associative-algebras magmas
asked 3 hours ago
John Baez
8,0384189
asked 3 hours ago
John Baez
8,0384189
asked 3 hours ago
John Baez
8,0384189
asked 3 hours ago
John Baez
8,0384189
asked 3 hours ago
John Baez
8,0384189
8,0384189
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I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19:
Suppose $Q(+)$ is an abelian group of even order $ngeq 6$. Let $a,bin Q-0$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $xnotin b,a+b$ or $ynotin b,a+b$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.
Then $Q(cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.
Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19:
Suppose $Q(+)$ is an abelian group of even order $ngeq 6$. Let $a,bin Q-0$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $xnotin b,a+b$ or $ynotin b,a+b$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.
Then $Q(cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.
Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
add a comment |Â
up vote
3
down vote
I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19:
Suppose $Q(+)$ is an abelian group of even order $ngeq 6$. Let $a,bin Q-0$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $xnotin b,a+b$ or $ynotin b,a+b$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.
Then $Q(cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.
Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19:
Suppose $Q(+)$ is an abelian group of even order $ngeq 6$. Let $a,bin Q-0$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $xnotin b,a+b$ or $ynotin b,a+b$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.
Then $Q(cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.
Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19:
Suppose $Q(+)$ is an abelian group of even order $ngeq 6$. Let $a,bin Q-0$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $xnotin b,a+b$ or $ynotin b,a+b$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.
Then $Q(cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.
Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
edited 2 hours ago
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
answered 2 hours ago
Gjergji Zaimi
59.1k3153293
59.1k3153293
add a comment |Â
add a comment |Â
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