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Is the free group on two generators generated by two elements?
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$begingroup$
...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 2 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 2 hours ago
MartianInvaderMartianInvader
5,2631325
$endgroup$
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
46 mins ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
38 mins 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 2 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 2 hours ago
MartianInvaderMartianInvader
5,2631325
asked 2 hours ago
MartianInvaderMartianInvader
5,2631325
asked 2 hours ago
MartianInvaderMartianInvader
5,2631325
asked 2 hours ago
MartianInvaderMartianInvader
5,2631325
asked 2 hours ago
MartianInvaderMartianInvader
5,2631325
5,2631325
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
46 mins ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
38 mins ago
add a comment |
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
46 mins ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
38 mins ago
1
1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
46 mins ago
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
46 mins ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
38 mins ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
38 mins 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 1 hour 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
20 mins 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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oldest
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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 1 hour 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
20 mins 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 1 hour 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
20 mins 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 1 hour 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 1 hour ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
edited 1 hour ago
answered 1 hour ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
answered 1 hour ago
Lord Shark the UnknownLord Shark the Unknown
111k1164139
answered 1 hour 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
20 mins 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
20 mins ago
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
20 mins ago
$begingroup$
Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning.
$endgroup$
– MartianInvader
20 mins ago
add a comment |
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1
$begingroup$
Nice fake. You had me going there.
$endgroup$
– Shalop
46 mins ago
$begingroup$
You might ask the same question about the free group on one generator.
$endgroup$
– Somos
38 mins ago