Calculating a presentation of $mathbb{Z}_{3}$ in detail.












1












$begingroup$



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







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$endgroup$








  • 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


















1












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 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
















1












1








1





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 4:35









Shaun

9,083113683




9,083113683










asked Dec 5 '18 at 1:34









eraldcoileraldcoil

395211




395211








  • 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




    $begingroup$
    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












1 Answer
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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 :)






share|cite|improve this answer











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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 :)






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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 :)






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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 :)






        share|cite|improve this answer











        $endgroup$



        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 :)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 3:34

























        answered Dec 14 '18 at 3:15









        ShaunShaun

        9,083113683




        9,083113683






























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