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Is the free group on two generators generated by two elements?
Group isomorphism concerning free group generated by $3$ elements.Why is chosen for intersection instead of union?Finitely generated ordered monoids and noetherian subsetsInjectivity of Natural Homomorphism to GroupificationCommutator Subgroup of Free GroupHow can you use Green's relations to learn about a monoid?Subgroup of the free group on 3 generatorsFinitely Generated Free Group to Finitely Generated Free MonoidIs there a generalization of the free group that includes infinitely long words?Show that the free group on three generators is a subgroup of the free group on two generators
$begingroup$
...as a monoid?
I was musing about this today and couldn't come up with an answer. Obviously it can be generated as a monoid by the four elements $a$, $b$, $a^-1$, and $b^-1$. After some playing around I was able to come up with three elements that generate it as a monoid: $ab$, $ab^-1$, and $a^-1$.
But I haven't been able to come up with two generators, nor an argument as to why that should be impossible.
free-groups monoid
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
$endgroup$
add a comment |
$begingroup$
...as a monoid?
I was musing about this today and couldn't come up with an answer. Obviously it can be generated as a monoid by the four elements $a$, $b$, $a^-1$, and $b^-1$. After some playing around I was able to come up with three elements that generate it as a monoid: $ab$, $ab^-1$, and $a^-1$.
But I haven't been able to come up with two generators, nor an argument as to why that should be impossible.
free-groups monoid
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
$endgroup$
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
2 hours ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
2 hours ago
add a comment |
$begingroup$
...as a monoid?
I was musing about this today and couldn't come up with an answer. Obviously it can be generated as a monoid by the four elements $a$, $b$, $a^-1$, and $b^-1$. After some playing around I was able to come up with three elements that generate it as a monoid: $ab$, $ab^-1$, and $a^-1$.
But I haven't been able to come up with two generators, nor an argument as to why that should be impossible.
free-groups monoid
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
$endgroup$
...as a monoid?
I was musing about this today and couldn't come up with an answer. Obviously it can be generated as a monoid by the four elements $a$, $b$, $a^-1$, and $b^-1$. After some playing around I was able to come up with three elements that generate it as a monoid: $ab$, $ab^-1$, and $a^-1$.
But I haven't been able to come up with two generators, nor an argument as to why that should be impossible.
free-groups monoid
free-groups monoid
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
asked 3 hours ago
MartianInvaderMartianInvader
5,2631325
5,2631325
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
2 hours ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
2 hours ago
add a comment |
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
2 hours ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
2 hours ago
1
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
2 hours ago
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
2 hours ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
2 hours ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The free group on two generators maps onto $Bbb Z^2$. (This is its Abelianisation). If it were generated by two elements as a monoid, then so
would $Bbb Z^2$. But that's not so.
If you have two elements $a$, $b$ of $Bbb Z^2$
generating it as a monoid, they certainly generate it as an Abelian group,
so they must be linearly independent as vectors. But in that case
$-a-b$ is not in the submonoid of $Bbb Z^2$ generated by $a$ and $b$.
Likewise, a free group on $n$ generators cannot be generated as a monoid
by $n$ elements.
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
$endgroup$
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
2 hours ago
add a comment |
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1 Answer
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1 Answer
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active
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active
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votes
$begingroup$
The free group on two generators maps onto $Bbb Z^2$. (This is its Abelianisation). If it were generated by two elements as a monoid, then so
would $Bbb Z^2$. But that's not so.
If you have two elements $a$, $b$ of $Bbb Z^2$
generating it as a monoid, they certainly generate it as an Abelian group,
so they must be linearly independent as vectors. But in that case
$-a-b$ is not in the submonoid of $Bbb Z^2$ generated by $a$ and $b$.
Likewise, a free group on $n$ generators cannot be generated as a monoid
by $n$ elements.
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
$endgroup$
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
2 hours ago
add a comment |
$begingroup$
The free group on two generators maps onto $Bbb Z^2$. (This is its Abelianisation). If it were generated by two elements as a monoid, then so
would $Bbb Z^2$. But that's not so.
If you have two elements $a$, $b$ of $Bbb Z^2$
generating it as a monoid, they certainly generate it as an Abelian group,
so they must be linearly independent as vectors. But in that case
$-a-b$ is not in the submonoid of $Bbb Z^2$ generated by $a$ and $b$.
Likewise, a free group on $n$ generators cannot be generated as a monoid
by $n$ elements.
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
$endgroup$
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
2 hours ago
add a comment |
$begingroup$
The free group on two generators maps onto $Bbb Z^2$. (This is its Abelianisation). If it were generated by two elements as a monoid, then so
would $Bbb Z^2$. But that's not so.
If you have two elements $a$, $b$ of $Bbb Z^2$
generating it as a monoid, they certainly generate it as an Abelian group,
so they must be linearly independent as vectors. But in that case
$-a-b$ is not in the submonoid of $Bbb Z^2$ generated by $a$ and $b$.
Likewise, a free group on $n$ generators cannot be generated as a monoid
by $n$ elements.
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
$endgroup$
The free group on two generators maps onto $Bbb Z^2$. (This is its Abelianisation). If it were generated by two elements as a monoid, then so
would $Bbb Z^2$. But that's not so.
If you have two elements $a$, $b$ of $Bbb Z^2$
generating it as a monoid, they certainly generate it as an Abelian group,
so they must be linearly independent as vectors. But in that case
$-a-b$ is not in the submonoid of $Bbb Z^2$ generated by $a$ and $b$.
Likewise, a free group on $n$ generators cannot be generated as a monoid
by $n$ elements.
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
edited 3 hours ago
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
answered 3 hours ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
111k1164139
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
2 hours ago
add a comment |
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
2 hours ago
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
2 hours ago
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
2 hours ago
add a comment |
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1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
2 hours ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
2 hours ago