Calculating a presentation of $mathbb{Z}_{3}$ in detail.
$begingroup$
Theorem: Let $G$ groups and $Ssubset G$ such that
$langle Srangle =G$. (Here $G=left{s_1ldots, s_n:s_iin Scup S^{-1}, ninmathbb{N}right}$.) Let $varphi:Sto G$ with $varphi(s)=s$. By the universal property of free groups there exists a unique homomorphism (in fact, epimorphism) $varphi:F(S)to G$ with $$F(S)=left{win S^{ast}: w text{ reduced word} right}.$$Then $$Gsimeq F(S)/{ker(varphi)}.$$
Here $langle langle Sranglerangle=langle left{gsg^{-1}:sin Scup S^{-1}, gin Gright}rangle.$
Let $Ssubset G$ and $G=langle Srangle.$ Then $langle Smid Trangle $ presentation of $G$ if $G=langle Srangle$ and $Tsubset kervarphi$ and $langle langle Tranglerangle=kervarphi$.
I want prove in a detailed way that $mathbb{Z}_{3}=langle amid a^3rangle.$
I have this.
Here $S=left{aright}.$
First, $a$ must be different from $0$. Because if $a=0$, then $mathbb{Z}_{3}=left{0right}$.
If $a=1,$ then $0=1+1+1, 1=1, 2=1+1$.
If $a=2$, then $0=2+2+2, 1=2+2^{-1}, 2=2$.
Therefore, $mathbb{Z}_{3}=langle arangle$, with $a=1$ or $a=2$.
So . . .
How prove $ker(varphi)=langlelangle left{a^3right}ranglerangle$?
I have this:
$kervarphi=left{s_1cdots s_nin F(S): varphi(s_1cdots s_n)=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
$=left{s_1cdots s_nin F(S): s_1+cdots +s_n=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
group-theory finite-groups cyclic-groups group-presentation universal-property
edited Dec 14 '18 at 4:35
Shaun
8,951113682
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
$endgroup$
add a comment |
$begingroup$
Theorem: Let $G$ groups and $Ssubset G$ such that
$langle Srangle =G$. (Here $G=left{s_1ldots, s_n:s_iin Scup S^{-1}, ninmathbb{N}right}$.) Let $varphi:Sto G$ with $varphi(s)=s$. By the universal property of free groups there exists a unique homomorphism (in fact, epimorphism) $varphi:F(S)to G$ with $$F(S)=left{win S^{ast}: w text{ reduced word} right}.$$Then $$Gsimeq F(S)/{ker(varphi)}.$$
Here $langle langle Sranglerangle=langle left{gsg^{-1}:sin Scup S^{-1}, gin Gright}rangle.$
Let $Ssubset G$ and $G=langle Srangle.$ Then $langle Smid Trangle $ presentation of $G$ if $G=langle Srangle$ and $Tsubset kervarphi$ and $langle langle Tranglerangle=kervarphi$.
I want prove in a detailed way that $mathbb{Z}_{3}=langle amid a^3rangle.$
I have this.
Here $S=left{aright}.$
First, $a$ must be different from $0$. Because if $a=0$, then $mathbb{Z}_{3}=left{0right}$.
If $a=1,$ then $0=1+1+1, 1=1, 2=1+1$.
If $a=2$, then $0=2+2+2, 1=2+2^{-1}, 2=2$.
Therefore, $mathbb{Z}_{3}=langle arangle$, with $a=1$ or $a=2$.
So . . .
How prove $ker(varphi)=langlelangle left{a^3right}ranglerangle$?
I have this:
$kervarphi=left{s_1cdots s_nin F(S): varphi(s_1cdots s_n)=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
$=left{s_1cdots s_nin F(S): s_1+cdots +s_n=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
group-theory finite-groups cyclic-groups group-presentation universal-property
edited Dec 14 '18 at 4:35
Shaun
8,951113682
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
$endgroup$
2
$begingroup$
There are many different ways of defining $Bbb Z_3$ and a sufficiently detailed answer would depend on which definition you are using. Please be more specific in future.
$endgroup$
– Shaun
Dec 14 '18 at 3:32
1
$begingroup$
Note, too, that there is no such thing as the presentation of a group, strictly speaking.
$endgroup$
– Shaun
Dec 14 '18 at 4:36
add a comment |
$begingroup$
Theorem: Let $G$ groups and $Ssubset G$ such that
$langle Srangle =G$. (Here $G=left{s_1ldots, s_n:s_iin Scup S^{-1}, ninmathbb{N}right}$.) Let $varphi:Sto G$ with $varphi(s)=s$. By the universal property of free groups there exists a unique homomorphism (in fact, epimorphism) $varphi:F(S)to G$ with $$F(S)=left{win S^{ast}: w text{ reduced word} right}.$$Then $$Gsimeq F(S)/{ker(varphi)}.$$
Here $langle langle Sranglerangle=langle left{gsg^{-1}:sin Scup S^{-1}, gin Gright}rangle.$
Let $Ssubset G$ and $G=langle Srangle.$ Then $langle Smid Trangle $ presentation of $G$ if $G=langle Srangle$ and $Tsubset kervarphi$ and $langle langle Tranglerangle=kervarphi$.
I want prove in a detailed way that $mathbb{Z}_{3}=langle amid a^3rangle.$
I have this.
Here $S=left{aright}.$
First, $a$ must be different from $0$. Because if $a=0$, then $mathbb{Z}_{3}=left{0right}$.
If $a=1,$ then $0=1+1+1, 1=1, 2=1+1$.
If $a=2$, then $0=2+2+2, 1=2+2^{-1}, 2=2$.
Therefore, $mathbb{Z}_{3}=langle arangle$, with $a=1$ or $a=2$.
So . . .
How prove $ker(varphi)=langlelangle left{a^3right}ranglerangle$?
I have this:
$kervarphi=left{s_1cdots s_nin F(S): varphi(s_1cdots s_n)=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
$=left{s_1cdots s_nin F(S): s_1+cdots +s_n=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
group-theory finite-groups cyclic-groups group-presentation universal-property
edited Dec 14 '18 at 4:35
Shaun
8,951113682
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
$endgroup$
Theorem: Let $G$ groups and $Ssubset G$ such that
$langle Srangle =G$. (Here $G=left{s_1ldots, s_n:s_iin Scup S^{-1}, ninmathbb{N}right}$.) Let $varphi:Sto G$ with $varphi(s)=s$. By the universal property of free groups there exists a unique homomorphism (in fact, epimorphism) $varphi:F(S)to G$ with $$F(S)=left{win S^{ast}: w text{ reduced word} right}.$$Then $$Gsimeq F(S)/{ker(varphi)}.$$
Here $langle langle Sranglerangle=langle left{gsg^{-1}:sin Scup S^{-1}, gin Gright}rangle.$
Let $Ssubset G$ and $G=langle Srangle.$ Then $langle Smid Trangle $ presentation of $G$ if $G=langle Srangle$ and $Tsubset kervarphi$ and $langle langle Tranglerangle=kervarphi$.
I want prove in a detailed way that $mathbb{Z}_{3}=langle amid a^3rangle.$
I have this.
Here $S=left{aright}.$
First, $a$ must be different from $0$. Because if $a=0$, then $mathbb{Z}_{3}=left{0right}$.
If $a=1,$ then $0=1+1+1, 1=1, 2=1+1$.
If $a=2$, then $0=2+2+2, 1=2+2^{-1}, 2=2$.
Therefore, $mathbb{Z}_{3}=langle arangle$, with $a=1$ or $a=2$.
So . . .
How prove $ker(varphi)=langlelangle left{a^3right}ranglerangle$?
I have this:
$kervarphi=left{s_1cdots s_nin F(S): varphi(s_1cdots s_n)=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
$=left{s_1cdots s_nin F(S): s_1+cdots +s_n=0, s_iin left{aright}cupleft{a^{-1}right}, ninmathbb{N}right}$
group-theory finite-groups cyclic-groups group-presentation universal-property
group-theory finite-groups cyclic-groups group-presentation universal-property
edited Dec 14 '18 at 4:35
Shaun
8,951113682
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
edited Dec 14 '18 at 4:35
Shaun
8,951113682
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
edited Dec 14 '18 at 4:35
Shaun
8,951113682
edited Dec 14 '18 at 4:35
Shaun
8,951113682
edited Dec 14 '18 at 4:35
Shaun
8,951113682
8,951113682
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
asked Dec 5 '18 at 1:34
eraldcoileraldcoil
385211
385211
2
$begingroup$
There are many different ways of defining $Bbb Z_3$ and a sufficiently detailed answer would depend on which definition you are using. Please be more specific in future.
$endgroup$
– Shaun
Dec 14 '18 at 3:32
1
$begingroup$
Note, too, that there is no such thing as the presentation of a group, strictly speaking.
$endgroup$
– Shaun
Dec 14 '18 at 4:36
add a comment |
2
$begingroup$
There are many different ways of defining $Bbb Z_3$ and a sufficiently detailed answer would depend on which definition you are using. Please be more specific in future.
$endgroup$
– Shaun
Dec 14 '18 at 3:32
1
$begingroup$
Note, too, that there is no such thing as the presentation of a group, strictly speaking.
$endgroup$
– Shaun
Dec 14 '18 at 4:36
2
2
$begingroup$
There are many different ways of defining $Bbb Z_3$ and a sufficiently detailed answer would depend on which definition you are using. Please be more specific in future.
$endgroup$
– Shaun
Dec 14 '18 at 3:32
$begingroup$
There are many different ways of defining $Bbb Z_3$ and a sufficiently detailed answer would depend on which definition you are using. Please be more specific in future.
$endgroup$
– Shaun
Dec 14 '18 at 3:32
1
1
$begingroup$
Note, too, that there is no such thing as the presentation of a group, strictly speaking.
$endgroup$
– Shaun
Dec 14 '18 at 4:36
$begingroup$
Note, too, that there is no such thing as the presentation of a group, strictly speaking.
$endgroup$
– Shaun
Dec 14 '18 at 4:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In order to show $ker(varphi)=langle langle a^3ranglerangle$, one must first be clear on what $varphi$ is, and instead of writing, say, $color{red}{a=1}$, one writes $varphi(a)=1$.
You are on the right track in that you have evidence to suggest that $varphi: amapsto 1text{ or } 2$; that is, that the generator $a$ of the presentation $langle amid a^3rangle$ maps via $varphi$ to one of the elements of $Bbb Z_3$ given by a number coprime to $3$.
Ask yourself,
What reduced words, with letters in ${ a, a^{-1}}$, get mapped to the identity of $Bbb Z_3$ via $varphi$?
But can you deduce what each word maps to under $varphi$ in general?
Hover your cursor over (or, if you're on a touchscreen device, tap) the box below for some hints.
Hint: Use Bézout's Identity, assuming $varphi(a)=p$ for some $p$ coprime to $3$. Here $varphi$ is a homomorphism. Don't forget to show both $ker(varphi)subseteq langlelangle a^3ranglerangle$ and $langlelangle a^3rangleranglesubseteqker(varphi)$.
I hope this helps :)
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
$endgroup$
add a comment |
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$begingroup$
In order to show $ker(varphi)=langle langle a^3ranglerangle$, one must first be clear on what $varphi$ is, and instead of writing, say, $color{red}{a=1}$, one writes $varphi(a)=1$.
You are on the right track in that you have evidence to suggest that $varphi: amapsto 1text{ or } 2$; that is, that the generator $a$ of the presentation $langle amid a^3rangle$ maps via $varphi$ to one of the elements of $Bbb Z_3$ given by a number coprime to $3$.
Ask yourself,
What reduced words, with letters in ${ a, a^{-1}}$, get mapped to the identity of $Bbb Z_3$ via $varphi$?
But can you deduce what each word maps to under $varphi$ in general?
Hover your cursor over (or, if you're on a touchscreen device, tap) the box below for some hints.
Hint: Use Bézout's Identity, assuming $varphi(a)=p$ for some $p$ coprime to $3$. Here $varphi$ is a homomorphism. Don't forget to show both $ker(varphi)subseteq langlelangle a^3ranglerangle$ and $langlelangle a^3rangleranglesubseteqker(varphi)$.
I hope this helps :)
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
$endgroup$
add a comment |
$begingroup$
In order to show $ker(varphi)=langle langle a^3ranglerangle$, one must first be clear on what $varphi$ is, and instead of writing, say, $color{red}{a=1}$, one writes $varphi(a)=1$.
You are on the right track in that you have evidence to suggest that $varphi: amapsto 1text{ or } 2$; that is, that the generator $a$ of the presentation $langle amid a^3rangle$ maps via $varphi$ to one of the elements of $Bbb Z_3$ given by a number coprime to $3$.
Ask yourself,
What reduced words, with letters in ${ a, a^{-1}}$, get mapped to the identity of $Bbb Z_3$ via $varphi$?
But can you deduce what each word maps to under $varphi$ in general?
Hover your cursor over (or, if you're on a touchscreen device, tap) the box below for some hints.
Hint: Use Bézout's Identity, assuming $varphi(a)=p$ for some $p$ coprime to $3$. Here $varphi$ is a homomorphism. Don't forget to show both $ker(varphi)subseteq langlelangle a^3ranglerangle$ and $langlelangle a^3rangleranglesubseteqker(varphi)$.
I hope this helps :)
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
$endgroup$
add a comment |
$begingroup$
In order to show $ker(varphi)=langle langle a^3ranglerangle$, one must first be clear on what $varphi$ is, and instead of writing, say, $color{red}{a=1}$, one writes $varphi(a)=1$.
You are on the right track in that you have evidence to suggest that $varphi: amapsto 1text{ or } 2$; that is, that the generator $a$ of the presentation $langle amid a^3rangle$ maps via $varphi$ to one of the elements of $Bbb Z_3$ given by a number coprime to $3$.
Ask yourself,
What reduced words, with letters in ${ a, a^{-1}}$, get mapped to the identity of $Bbb Z_3$ via $varphi$?
But can you deduce what each word maps to under $varphi$ in general?
Hover your cursor over (or, if you're on a touchscreen device, tap) the box below for some hints.
Hint: Use Bézout's Identity, assuming $varphi(a)=p$ for some $p$ coprime to $3$. Here $varphi$ is a homomorphism. Don't forget to show both $ker(varphi)subseteq langlelangle a^3ranglerangle$ and $langlelangle a^3rangleranglesubseteqker(varphi)$.
I hope this helps :)
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
$endgroup$
In order to show $ker(varphi)=langle langle a^3ranglerangle$, one must first be clear on what $varphi$ is, and instead of writing, say, $color{red}{a=1}$, one writes $varphi(a)=1$.
You are on the right track in that you have evidence to suggest that $varphi: amapsto 1text{ or } 2$; that is, that the generator $a$ of the presentation $langle amid a^3rangle$ maps via $varphi$ to one of the elements of $Bbb Z_3$ given by a number coprime to $3$.
Ask yourself,
What reduced words, with letters in ${ a, a^{-1}}$, get mapped to the identity of $Bbb Z_3$ via $varphi$?
But can you deduce what each word maps to under $varphi$ in general?
Hover your cursor over (or, if you're on a touchscreen device, tap) the box below for some hints.
Hint: Use Bézout's Identity, assuming $varphi(a)=p$ for some $p$ coprime to $3$. Here $varphi$ is a homomorphism. Don't forget to show both $ker(varphi)subseteq langlelangle a^3ranglerangle$ and $langlelangle a^3rangleranglesubseteqker(varphi)$.
I hope this helps :)
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
edited Dec 14 '18 at 3:34
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
answered Dec 14 '18 at 3:15
ShaunShaun
8,951113682
8,951113682
add a comment |
add a comment |
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$begingroup$
There are many different ways of defining $Bbb Z_3$ and a sufficiently detailed answer would depend on which definition you are using. Please be more specific in future.
$endgroup$
– Shaun
Dec 14 '18 at 3:32
1
$begingroup$
Note, too, that there is no such thing as the presentation of a group, strictly speaking.
$endgroup$
– Shaun
Dec 14 '18 at 4:36