Mixing
How many generators do a 4-strands braid group have?
$begingroup$
I know that this might be a trivial question, but, I'm asking that because source on the topic tend to give different presentations.
Maybe it's just me not noticing the difference, but I'd like to point it out bacuase I think it can be helpful to avoid confusion for someone who is approaching braid groups. The problem lies in what is considered generators of abraid group, in particular a 4-strands to be specific. Are inverses of the 3 "standard" braids included in the set of generators?($displaystyle sigma 1$, $displaystyle sigma 2$, $displaystyle sigma 3$).
For example Wikipedia article talks about 3 generators while a bunch of other sources include the inverses ($displaystyle sigma 1^{-1}$, $displaystyle sigma 2^{-1}$, $displaystyle sigma 3^{-1}$) in the set of generators.
My intuition tells me to include them too, but I'm afraid I'm missing ssomething.
group-theory braid-groups
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
$endgroup$
|
show 5 more comments
$begingroup$
I know that this might be a trivial question, but, I'm asking that because source on the topic tend to give different presentations.
Maybe it's just me not noticing the difference, but I'd like to point it out bacuase I think it can be helpful to avoid confusion for someone who is approaching braid groups. The problem lies in what is considered generators of abraid group, in particular a 4-strands to be specific. Are inverses of the 3 "standard" braids included in the set of generators?($displaystyle sigma 1$, $displaystyle sigma 2$, $displaystyle sigma 3$).
For example Wikipedia article talks about 3 generators while a bunch of other sources include the inverses ($displaystyle sigma 1^{-1}$, $displaystyle sigma 2^{-1}$, $displaystyle sigma 3^{-1}$) in the set of generators.
My intuition tells me to include them too, but I'm afraid I'm missing ssomething.
group-theory braid-groups
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
$endgroup$
$begingroup$
Inverses are not needed as they appear automatically: A subset $S$ of a group $G$ generates $G$, so $G=langle Srangle$, if no proper subgroup of $G$ contains the set $S$. The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses. [For finite groups, $langle Srangle$ may be views as the product of all elements of $S$.] For example, $mathbb{Z}=langle 1rangle$ even though $-3$ cannot be written as a sum of $1$s.
$endgroup$
– user1729
Jan 24 at 14:41
$begingroup$
What do you mean you "want to show that there is always confusion"?
$endgroup$
– Tobias Kildetoft
Jan 24 at 14:42
$begingroup$
@TobiasKildetoft I editeted that part because it was misleading about what I meant! :). I just wanted to say that it's easy to find different presentations.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:09
$begingroup$
@user1729 I didn't get what you mean, maybe it's too advanced for me.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:12
$begingroup$
Its not too advanced, don't worry :-) First ask yourself the following question: does $1$ generate $mathbb{Z}$?
$endgroup$
– user1729
Jan 24 at 15:14
|
show 5 more comments
$begingroup$
I know that this might be a trivial question, but, I'm asking that because source on the topic tend to give different presentations.
Maybe it's just me not noticing the difference, but I'd like to point it out bacuase I think it can be helpful to avoid confusion for someone who is approaching braid groups. The problem lies in what is considered generators of abraid group, in particular a 4-strands to be specific. Are inverses of the 3 "standard" braids included in the set of generators?($displaystyle sigma 1$, $displaystyle sigma 2$, $displaystyle sigma 3$).
For example Wikipedia article talks about 3 generators while a bunch of other sources include the inverses ($displaystyle sigma 1^{-1}$, $displaystyle sigma 2^{-1}$, $displaystyle sigma 3^{-1}$) in the set of generators.
My intuition tells me to include them too, but I'm afraid I'm missing ssomething.
group-theory braid-groups
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
$endgroup$
I know that this might be a trivial question, but, I'm asking that because source on the topic tend to give different presentations.
Maybe it's just me not noticing the difference, but I'd like to point it out bacuase I think it can be helpful to avoid confusion for someone who is approaching braid groups. The problem lies in what is considered generators of abraid group, in particular a 4-strands to be specific. Are inverses of the 3 "standard" braids included in the set of generators?($displaystyle sigma 1$, $displaystyle sigma 2$, $displaystyle sigma 3$).
For example Wikipedia article talks about 3 generators while a bunch of other sources include the inverses ($displaystyle sigma 1^{-1}$, $displaystyle sigma 2^{-1}$, $displaystyle sigma 3^{-1}$) in the set of generators.
My intuition tells me to include them too, but I'm afraid I'm missing ssomething.
group-theory braid-groups
group-theory braid-groups
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
edited Jan 24 at 15:08
Gabriele Scarlatti
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
asked Jan 24 at 14:33
Gabriele ScarlattiGabriele Scarlatti
370212
370212
$begingroup$
Inverses are not needed as they appear automatically: A subset $S$ of a group $G$ generates $G$, so $G=langle Srangle$, if no proper subgroup of $G$ contains the set $S$. The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses. [For finite groups, $langle Srangle$ may be views as the product of all elements of $S$.] For example, $mathbb{Z}=langle 1rangle$ even though $-3$ cannot be written as a sum of $1$s.
$endgroup$
– user1729
Jan 24 at 14:41
$begingroup$
What do you mean you "want to show that there is always confusion"?
$endgroup$
– Tobias Kildetoft
Jan 24 at 14:42
$begingroup$
@TobiasKildetoft I editeted that part because it was misleading about what I meant! :). I just wanted to say that it's easy to find different presentations.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:09
$begingroup$
@user1729 I didn't get what you mean, maybe it's too advanced for me.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:12
$begingroup$
Its not too advanced, don't worry :-) First ask yourself the following question: does $1$ generate $mathbb{Z}$?
$endgroup$
– user1729
Jan 24 at 15:14
|
show 5 more comments
$begingroup$
Inverses are not needed as they appear automatically: A subset $S$ of a group $G$ generates $G$, so $G=langle Srangle$, if no proper subgroup of $G$ contains the set $S$. The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses. [For finite groups, $langle Srangle$ may be views as the product of all elements of $S$.] For example, $mathbb{Z}=langle 1rangle$ even though $-3$ cannot be written as a sum of $1$s.
$endgroup$
– user1729
Jan 24 at 14:41
$begingroup$
What do you mean you "want to show that there is always confusion"?
$endgroup$
– Tobias Kildetoft
Jan 24 at 14:42
$begingroup$
@TobiasKildetoft I editeted that part because it was misleading about what I meant! :). I just wanted to say that it's easy to find different presentations.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:09
$begingroup$
@user1729 I didn't get what you mean, maybe it's too advanced for me.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:12
$begingroup$
Its not too advanced, don't worry :-) First ask yourself the following question: does $1$ generate $mathbb{Z}$?
$endgroup$
– user1729
Jan 24 at 15:14
$begingroup$
Inverses are not needed as they appear automatically: A subset $S$ of a group $G$ generates $G$, so $G=langle Srangle$, if no proper subgroup of $G$ contains the set $S$. The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses. [For finite groups, $langle Srangle$ may be views as the product of all elements of $S$.] For example, $mathbb{Z}=langle 1rangle$ even though $-3$ cannot be written as a sum of $1$s.
$endgroup$
– user1729
Jan 24 at 14:41
$begingroup$
Inverses are not needed as they appear automatically: A subset $S$ of a group $G$ generates $G$, so $G=langle Srangle$, if no proper subgroup of $G$ contains the set $S$. The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses. [For finite groups, $langle Srangle$ may be views as the product of all elements of $S$.] For example, $mathbb{Z}=langle 1rangle$ even though $-3$ cannot be written as a sum of $1$s.
$endgroup$
– user1729
Jan 24 at 14:41
$begingroup$
What do you mean you "want to show that there is always confusion"?
$endgroup$
– Tobias Kildetoft
Jan 24 at 14:42
$begingroup$
What do you mean you "want to show that there is always confusion"?
$endgroup$
– Tobias Kildetoft
Jan 24 at 14:42
$begingroup$
@TobiasKildetoft I editeted that part because it was misleading about what I meant! :). I just wanted to say that it's easy to find different presentations.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:09
$begingroup$
@TobiasKildetoft I editeted that part because it was misleading about what I meant! :). I just wanted to say that it's easy to find different presentations.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:09
$begingroup$
@user1729 I didn't get what you mean, maybe it's too advanced for me.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:12
$begingroup$
@user1729 I didn't get what you mean, maybe it's too advanced for me.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:12
$begingroup$
Its not too advanced, don't worry :-) First ask yourself the following question: does $1$ generate $mathbb{Z}$?
$endgroup$
– user1729
Jan 24 at 15:14
$begingroup$
Its not too advanced, don't worry :-) First ask yourself the following question: does $1$ generate $mathbb{Z}$?
$endgroup$
– user1729
Jan 24 at 15:14
|
show 5 more comments
1 Answer
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$begingroup$
Definition. Let $Ssubset G$. Then the subgroup generated by $S$, written $langle Srangle$, is the smallest subgroup of $G$ containing $S$. Formally: $langle Srangle:=displaystylebigcap_{Ssubset H,\ Hleq G}H$.
Then a generating set of $G$ is a subset $S$ of $G$ where $langle Srangle=G$. For example, $langle 1rangle=mathbb{Z}$.
There is nothing in this definition about "products of elements". However, it turns out that the following holds:
A set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $Scup S^{-1}$ (where $S^{-1}:={s^{-1}mid sin S}$ is the set of inverses of elements of $S$).
The phrase from Wikipedia "The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses" is implying that the inverses of $S$ are not necessarily in $S$! For example, every integer can be written as a sum of $1$s and of $-1$s, so $langle 1rangle=mathbb{Z}$. To be clear: there is no assumption that $S=S^{-1}$.
Lets end on an exercise, which basically says you can ignore this subtlety for finite groups:
Exercise. Prove that set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $S$.
answered Jan 24 at 16:37
user1729user1729
17.4k64193
$endgroup$
add a comment |
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$begingroup$
Definition. Let $Ssubset G$. Then the subgroup generated by $S$, written $langle Srangle$, is the smallest subgroup of $G$ containing $S$. Formally: $langle Srangle:=displaystylebigcap_{Ssubset H,\ Hleq G}H$.
Then a generating set of $G$ is a subset $S$ of $G$ where $langle Srangle=G$. For example, $langle 1rangle=mathbb{Z}$.
There is nothing in this definition about "products of elements". However, it turns out that the following holds:
A set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $Scup S^{-1}$ (where $S^{-1}:={s^{-1}mid sin S}$ is the set of inverses of elements of $S$).
The phrase from Wikipedia "The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses" is implying that the inverses of $S$ are not necessarily in $S$! For example, every integer can be written as a sum of $1$s and of $-1$s, so $langle 1rangle=mathbb{Z}$. To be clear: there is no assumption that $S=S^{-1}$.
Lets end on an exercise, which basically says you can ignore this subtlety for finite groups:
Exercise. Prove that set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $S$.
answered Jan 24 at 16:37
user1729user1729
17.4k64193
$endgroup$
add a comment |
$begingroup$
Definition. Let $Ssubset G$. Then the subgroup generated by $S$, written $langle Srangle$, is the smallest subgroup of $G$ containing $S$. Formally: $langle Srangle:=displaystylebigcap_{Ssubset H,\ Hleq G}H$.
Then a generating set of $G$ is a subset $S$ of $G$ where $langle Srangle=G$. For example, $langle 1rangle=mathbb{Z}$.
There is nothing in this definition about "products of elements". However, it turns out that the following holds:
A set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $Scup S^{-1}$ (where $S^{-1}:={s^{-1}mid sin S}$ is the set of inverses of elements of $S$).
The phrase from Wikipedia "The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses" is implying that the inverses of $S$ are not necessarily in $S$! For example, every integer can be written as a sum of $1$s and of $-1$s, so $langle 1rangle=mathbb{Z}$. To be clear: there is no assumption that $S=S^{-1}$.
Lets end on an exercise, which basically says you can ignore this subtlety for finite groups:
Exercise. Prove that set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $S$.
answered Jan 24 at 16:37
user1729user1729
17.4k64193
$endgroup$
add a comment |
$begingroup$
Definition. Let $Ssubset G$. Then the subgroup generated by $S$, written $langle Srangle$, is the smallest subgroup of $G$ containing $S$. Formally: $langle Srangle:=displaystylebigcap_{Ssubset H,\ Hleq G}H$.
Then a generating set of $G$ is a subset $S$ of $G$ where $langle Srangle=G$. For example, $langle 1rangle=mathbb{Z}$.
There is nothing in this definition about "products of elements". However, it turns out that the following holds:
A set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $Scup S^{-1}$ (where $S^{-1}:={s^{-1}mid sin S}$ is the set of inverses of elements of $S$).
The phrase from Wikipedia "The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses" is implying that the inverses of $S$ are not necessarily in $S$! For example, every integer can be written as a sum of $1$s and of $-1$s, so $langle 1rangle=mathbb{Z}$. To be clear: there is no assumption that $S=S^{-1}$.
Lets end on an exercise, which basically says you can ignore this subtlety for finite groups:
Exercise. Prove that set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $S$.
answered Jan 24 at 16:37
user1729user1729
17.4k64193
$endgroup$
Definition. Let $Ssubset G$. Then the subgroup generated by $S$, written $langle Srangle$, is the smallest subgroup of $G$ containing $S$. Formally: $langle Srangle:=displaystylebigcap_{Ssubset H,\ Hleq G}H$.
Then a generating set of $G$ is a subset $S$ of $G$ where $langle Srangle=G$. For example, $langle 1rangle=mathbb{Z}$.
There is nothing in this definition about "products of elements". However, it turns out that the following holds:
A set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $Scup S^{-1}$ (where $S^{-1}:={s^{-1}mid sin S}$ is the set of inverses of elements of $S$).
The phrase from Wikipedia "The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses" is implying that the inverses of $S$ are not necessarily in $S$! For example, every integer can be written as a sum of $1$s and of $-1$s, so $langle 1rangle=mathbb{Z}$. To be clear: there is no assumption that $S=S^{-1}$.
Lets end on an exercise, which basically says you can ignore this subtlety for finite groups:
Exercise. Prove that set $S$ generates $G$ if and only if every element $gin G$ can be written as a product of elements from $S$.
answered Jan 24 at 16:37
user1729user1729
17.4k64193
answered Jan 24 at 16:37
user1729user1729
17.4k64193
answered Jan 24 at 16:37
user1729user1729
17.4k64193
answered Jan 24 at 16:37
user1729user1729
17.4k64193
17.4k64193
add a comment |
add a comment |
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$begingroup$
Inverses are not needed as they appear automatically: A subset $S$ of a group $G$ generates $G$, so $G=langle Srangle$, if no proper subgroup of $G$ contains the set $S$. The group $langle Srangle$ may be viewed as the product of all elements of $S$ and their inverses. [For finite groups, $langle Srangle$ may be views as the product of all elements of $S$.] For example, $mathbb{Z}=langle 1rangle$ even though $-3$ cannot be written as a sum of $1$s.
$endgroup$
– user1729
Jan 24 at 14:41
$begingroup$
What do you mean you "want to show that there is always confusion"?
$endgroup$
– Tobias Kildetoft
Jan 24 at 14:42
$begingroup$
@TobiasKildetoft I editeted that part because it was misleading about what I meant! :). I just wanted to say that it's easy to find different presentations.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:09
$begingroup$
@user1729 I didn't get what you mean, maybe it's too advanced for me.
$endgroup$
– Gabriele Scarlatti
Jan 24 at 15:12
$begingroup$
Its not too advanced, don't worry :-) First ask yourself the following question: does $1$ generate $mathbb{Z}$?
$endgroup$
– user1729
Jan 24 at 15:14