Check whether $G$ is group or not












5












$begingroup$



Let $G={0,1,2}$ define $*$ on $G$ such that $a*b=|a-b|$. Check if $G$ is a group




Edit : As @egreg correctly pointed out if you draw the Cayley table then one can directly see it is not a group, I was making a mistake in drawing up cayley table.



My solution :
I can prove that $G$ is closed, has an identity ($0$), has an inverse (each element is inverse of itself).



What I am not sure is whether the group is associative under this binary operation, at the face of it $|a-|b-c||neq||a-b|-c||$ but if we check all possibilities then both do come out to be equal if $a,b,c in G$.



Any hint/help will be appreciated






abstract-algebra group-theory binary-operations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    @Peter: I suspect at the level of this OP (reputation=1), that the statement that a group of size $3$ must be isomorphic to $mathbb{Z}_3$ will be of little help.
    $endgroup$
    – David G. Stork
    Jun 13 '18 at 12:45






  • 1




    $begingroup$
    Associativity is always the hardest property to verify but the good news is you don't have very many to check. You need to verify that $a*(b*c)=(a*b)*c$ so each letter has 3 choices $0,1,2$ so you'd have 27 things to check there if you manually check each one :)
    $endgroup$
    – N8tron
    Jun 13 '18 at 12:47






  • 2




    $begingroup$
    You only need to find one combination of values for $a,b$ and $c$ for which associativity fails and you can infer that they cannot all be the same number and $0$ doesn't really change the expression so try combinations of $1$ and $2$.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    You may also use the fact that in a group each row/column of the Cayley table must be a permutation of the elements of $G$.
    $endgroup$
    – Andreas Caranti
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    OK, an easier way to look if we CAN have a group : Does every element appear in every row and in every column ? If not, we cannot have a group.
    $endgroup$
    – Peter
    Jun 13 '18 at 12:49
















5












$begingroup$



Let $G={0,1,2}$ define $*$ on $G$ such that $a*b=|a-b|$. Check if $G$ is a group




Edit : As @egreg correctly pointed out if you draw the Cayley table then one can directly see it is not a group, I was making a mistake in drawing up cayley table.



My solution :
I can prove that $G$ is closed, has an identity ($0$), has an inverse (each element is inverse of itself).



What I am not sure is whether the group is associative under this binary operation, at the face of it $|a-|b-c||neq||a-b|-c||$ but if we check all possibilities then both do come out to be equal if $a,b,c in G$.



Any hint/help will be appreciated






abstract-algebra group-theory binary-operations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    @Peter: I suspect at the level of this OP (reputation=1), that the statement that a group of size $3$ must be isomorphic to $mathbb{Z}_3$ will be of little help.
    $endgroup$
    – David G. Stork
    Jun 13 '18 at 12:45






  • 1




    $begingroup$
    Associativity is always the hardest property to verify but the good news is you don't have very many to check. You need to verify that $a*(b*c)=(a*b)*c$ so each letter has 3 choices $0,1,2$ so you'd have 27 things to check there if you manually check each one :)
    $endgroup$
    – N8tron
    Jun 13 '18 at 12:47






  • 2




    $begingroup$
    You only need to find one combination of values for $a,b$ and $c$ for which associativity fails and you can infer that they cannot all be the same number and $0$ doesn't really change the expression so try combinations of $1$ and $2$.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    You may also use the fact that in a group each row/column of the Cayley table must be a permutation of the elements of $G$.
    $endgroup$
    – Andreas Caranti
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    OK, an easier way to look if we CAN have a group : Does every element appear in every row and in every column ? If not, we cannot have a group.
    $endgroup$
    – Peter
    Jun 13 '18 at 12:49














5












5








5





$begingroup$



Let $G={0,1,2}$ define $*$ on $G$ such that $a*b=|a-b|$. Check if $G$ is a group




Edit : As @egreg correctly pointed out if you draw the Cayley table then one can directly see it is not a group, I was making a mistake in drawing up cayley table.



My solution :
I can prove that $G$ is closed, has an identity ($0$), has an inverse (each element is inverse of itself).



What I am not sure is whether the group is associative under this binary operation, at the face of it $|a-|b-c||neq||a-b|-c||$ but if we check all possibilities then both do come out to be equal if $a,b,c in G$.



Any hint/help will be appreciated






abstract-algebra group-theory binary-operations







share|cite|improve this question











$endgroup$





Let $G={0,1,2}$ define $*$ on $G$ such that $a*b=|a-b|$. Check if $G$ is a group




Edit : As @egreg correctly pointed out if you draw the Cayley table then one can directly see it is not a group, I was making a mistake in drawing up cayley table.



My solution :
I can prove that $G$ is closed, has an identity ($0$), has an inverse (each element is inverse of itself).



What I am not sure is whether the group is associative under this binary operation, at the face of it $|a-|b-c||neq||a-b|-c||$ but if we check all possibilities then both do come out to be equal if $a,b,c in G$.



Any hint/help will be appreciated






abstract-algebra group-theory binary-operations




abstract-algebra group-theory binary-operations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 16:27









José Carlos Santos

157k22126227




157k22126227










asked Jun 13 '18 at 12:41









kira0705kira0705

995




995








  • 2




    $begingroup$
    @Peter: I suspect at the level of this OP (reputation=1), that the statement that a group of size $3$ must be isomorphic to $mathbb{Z}_3$ will be of little help.
    $endgroup$
    – David G. Stork
    Jun 13 '18 at 12:45






  • 1




    $begingroup$
    Associativity is always the hardest property to verify but the good news is you don't have very many to check. You need to verify that $a*(b*c)=(a*b)*c$ so each letter has 3 choices $0,1,2$ so you'd have 27 things to check there if you manually check each one :)
    $endgroup$
    – N8tron
    Jun 13 '18 at 12:47






  • 2




    $begingroup$
    You only need to find one combination of values for $a,b$ and $c$ for which associativity fails and you can infer that they cannot all be the same number and $0$ doesn't really change the expression so try combinations of $1$ and $2$.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    You may also use the fact that in a group each row/column of the Cayley table must be a permutation of the elements of $G$.
    $endgroup$
    – Andreas Caranti
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    OK, an easier way to look if we CAN have a group : Does every element appear in every row and in every column ? If not, we cannot have a group.
    $endgroup$
    – Peter
    Jun 13 '18 at 12:49














  • 2




    $begingroup$
    @Peter: I suspect at the level of this OP (reputation=1), that the statement that a group of size $3$ must be isomorphic to $mathbb{Z}_3$ will be of little help.
    $endgroup$
    – David G. Stork
    Jun 13 '18 at 12:45






  • 1




    $begingroup$
    Associativity is always the hardest property to verify but the good news is you don't have very many to check. You need to verify that $a*(b*c)=(a*b)*c$ so each letter has 3 choices $0,1,2$ so you'd have 27 things to check there if you manually check each one :)
    $endgroup$
    – N8tron
    Jun 13 '18 at 12:47






  • 2




    $begingroup$
    You only need to find one combination of values for $a,b$ and $c$ for which associativity fails and you can infer that they cannot all be the same number and $0$ doesn't really change the expression so try combinations of $1$ and $2$.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    You may also use the fact that in a group each row/column of the Cayley table must be a permutation of the elements of $G$.
    $endgroup$
    – Andreas Caranti
    Jun 13 '18 at 12:48






  • 2




    $begingroup$
    OK, an easier way to look if we CAN have a group : Does every element appear in every row and in every column ? If not, we cannot have a group.
    $endgroup$
    – Peter
    Jun 13 '18 at 12:49








2




2




$begingroup$
@Peter: I suspect at the level of this OP (reputation=1), that the statement that a group of size $3$ must be isomorphic to $mathbb{Z}_3$ will be of little help.
$endgroup$
– David G. Stork
Jun 13 '18 at 12:45




$begingroup$
@Peter: I suspect at the level of this OP (reputation=1), that the statement that a group of size $3$ must be isomorphic to $mathbb{Z}_3$ will be of little help.
$endgroup$
– David G. Stork
Jun 13 '18 at 12:45




1




1




$begingroup$
Associativity is always the hardest property to verify but the good news is you don't have very many to check. You need to verify that $a*(b*c)=(a*b)*c$ so each letter has 3 choices $0,1,2$ so you'd have 27 things to check there if you manually check each one :)
$endgroup$
– N8tron
Jun 13 '18 at 12:47




$begingroup$
Associativity is always the hardest property to verify but the good news is you don't have very many to check. You need to verify that $a*(b*c)=(a*b)*c$ so each letter has 3 choices $0,1,2$ so you'd have 27 things to check there if you manually check each one :)
$endgroup$
– N8tron
Jun 13 '18 at 12:47




2




2




$begingroup$
You only need to find one combination of values for $a,b$ and $c$ for which associativity fails and you can infer that they cannot all be the same number and $0$ doesn't really change the expression so try combinations of $1$ and $2$.
$endgroup$
– CyclotomicField
Jun 13 '18 at 12:48




$begingroup$
You only need to find one combination of values for $a,b$ and $c$ for which associativity fails and you can infer that they cannot all be the same number and $0$ doesn't really change the expression so try combinations of $1$ and $2$.
$endgroup$
– CyclotomicField
Jun 13 '18 at 12:48




2




2




$begingroup$
You may also use the fact that in a group each row/column of the Cayley table must be a permutation of the elements of $G$.
$endgroup$
– Andreas Caranti
Jun 13 '18 at 12:48




$begingroup$
You may also use the fact that in a group each row/column of the Cayley table must be a permutation of the elements of $G$.
$endgroup$
– Andreas Caranti
Jun 13 '18 at 12:48




2




2




$begingroup$
OK, an easier way to look if we CAN have a group : Does every element appear in every row and in every column ? If not, we cannot have a group.
$endgroup$
– Peter
Jun 13 '18 at 12:49




$begingroup$
OK, an easier way to look if we CAN have a group : Does every element appear in every row and in every column ? If not, we cannot have a group.
$endgroup$
– Peter
Jun 13 '18 at 12:49










2 Answers
2






active

oldest

votes


















4












$begingroup$

The Cayley table (that also shows closure, identity and inverses) is
begin{array}{c|ccc}
* & 0 & 1 & 2 \
hline
0 & 0 & 1 & 2 \
1 & 1 & 0 & 1\
2 & 2 & 1 & 0
end{array}
Since in the second row the element $1$ appears twice, the set is not a group.



In a group, from $xy=xz$ one deduces $y=z$; here, instead, $1*0=1*2$, but $0ne2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I feel totally stupid , I even drew the cayley table wrong ! I will take care of such mistakes in future before posting any question , sorry for wasting your time .
    $endgroup$
    – kira0705
    Jun 13 '18 at 14:58






  • 2




    $begingroup$
    @kira0705 Don't worry: staring at some computation without realizing the error happens to everybody.
    $endgroup$
    – egreg
    Jun 13 '18 at 15:44



















9












$begingroup$

The operation $*$ is not associative, since:begin{align}bigl||2-1|-1bigr|&=0\&neq2\&=bigl|2-|1-1|bigr|.end{align}Therefore, $(G,*)$ is not a group.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Just wanted to add that associativity does work when $a=b=c$ or when any of the values is zero so to find a counter-example or prove associativity by brute force relatively few cases would need to be checked.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:52










  • $begingroup$
    thanks it perfectly answers my query !
    $endgroup$
    – kira0705
    Jun 13 '18 at 12:57











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The Cayley table (that also shows closure, identity and inverses) is
begin{array}{c|ccc}
* & 0 & 1 & 2 \
hline
0 & 0 & 1 & 2 \
1 & 1 & 0 & 1\
2 & 2 & 1 & 0
end{array}
Since in the second row the element $1$ appears twice, the set is not a group.



In a group, from $xy=xz$ one deduces $y=z$; here, instead, $1*0=1*2$, but $0ne2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I feel totally stupid , I even drew the cayley table wrong ! I will take care of such mistakes in future before posting any question , sorry for wasting your time .
    $endgroup$
    – kira0705
    Jun 13 '18 at 14:58






  • 2




    $begingroup$
    @kira0705 Don't worry: staring at some computation without realizing the error happens to everybody.
    $endgroup$
    – egreg
    Jun 13 '18 at 15:44
















4












$begingroup$

The Cayley table (that also shows closure, identity and inverses) is
begin{array}{c|ccc}
* & 0 & 1 & 2 \
hline
0 & 0 & 1 & 2 \
1 & 1 & 0 & 1\
2 & 2 & 1 & 0
end{array}
Since in the second row the element $1$ appears twice, the set is not a group.



In a group, from $xy=xz$ one deduces $y=z$; here, instead, $1*0=1*2$, but $0ne2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I feel totally stupid , I even drew the cayley table wrong ! I will take care of such mistakes in future before posting any question , sorry for wasting your time .
    $endgroup$
    – kira0705
    Jun 13 '18 at 14:58






  • 2




    $begingroup$
    @kira0705 Don't worry: staring at some computation without realizing the error happens to everybody.
    $endgroup$
    – egreg
    Jun 13 '18 at 15:44














4












4








4





$begingroup$

The Cayley table (that also shows closure, identity and inverses) is
begin{array}{c|ccc}
* & 0 & 1 & 2 \
hline
0 & 0 & 1 & 2 \
1 & 1 & 0 & 1\
2 & 2 & 1 & 0
end{array}
Since in the second row the element $1$ appears twice, the set is not a group.



In a group, from $xy=xz$ one deduces $y=z$; here, instead, $1*0=1*2$, but $0ne2$.






share|cite|improve this answer









$endgroup$



The Cayley table (that also shows closure, identity and inverses) is
begin{array}{c|ccc}
* & 0 & 1 & 2 \
hline
0 & 0 & 1 & 2 \
1 & 1 & 0 & 1\
2 & 2 & 1 & 0
end{array}
Since in the second row the element $1$ appears twice, the set is not a group.



In a group, from $xy=xz$ one deduces $y=z$; here, instead, $1*0=1*2$, but $0ne2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 13 '18 at 14:20









egregegreg

181k1485202




181k1485202












  • $begingroup$
    Thanks! I feel totally stupid , I even drew the cayley table wrong ! I will take care of such mistakes in future before posting any question , sorry for wasting your time .
    $endgroup$
    – kira0705
    Jun 13 '18 at 14:58






  • 2




    $begingroup$
    @kira0705 Don't worry: staring at some computation without realizing the error happens to everybody.
    $endgroup$
    – egreg
    Jun 13 '18 at 15:44


















  • $begingroup$
    Thanks! I feel totally stupid , I even drew the cayley table wrong ! I will take care of such mistakes in future before posting any question , sorry for wasting your time .
    $endgroup$
    – kira0705
    Jun 13 '18 at 14:58






  • 2




    $begingroup$
    @kira0705 Don't worry: staring at some computation without realizing the error happens to everybody.
    $endgroup$
    – egreg
    Jun 13 '18 at 15:44
















$begingroup$
Thanks! I feel totally stupid , I even drew the cayley table wrong ! I will take care of such mistakes in future before posting any question , sorry for wasting your time .
$endgroup$
– kira0705
Jun 13 '18 at 14:58




$begingroup$
Thanks! I feel totally stupid , I even drew the cayley table wrong ! I will take care of such mistakes in future before posting any question , sorry for wasting your time .
$endgroup$
– kira0705
Jun 13 '18 at 14:58




2




2




$begingroup$
@kira0705 Don't worry: staring at some computation without realizing the error happens to everybody.
$endgroup$
– egreg
Jun 13 '18 at 15:44




$begingroup$
@kira0705 Don't worry: staring at some computation without realizing the error happens to everybody.
$endgroup$
– egreg
Jun 13 '18 at 15:44











9












$begingroup$

The operation $*$ is not associative, since:begin{align}bigl||2-1|-1bigr|&=0\&neq2\&=bigl|2-|1-1|bigr|.end{align}Therefore, $(G,*)$ is not a group.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Just wanted to add that associativity does work when $a=b=c$ or when any of the values is zero so to find a counter-example or prove associativity by brute force relatively few cases would need to be checked.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:52










  • $begingroup$
    thanks it perfectly answers my query !
    $endgroup$
    – kira0705
    Jun 13 '18 at 12:57
















9












$begingroup$

The operation $*$ is not associative, since:begin{align}bigl||2-1|-1bigr|&=0\&neq2\&=bigl|2-|1-1|bigr|.end{align}Therefore, $(G,*)$ is not a group.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Just wanted to add that associativity does work when $a=b=c$ or when any of the values is zero so to find a counter-example or prove associativity by brute force relatively few cases would need to be checked.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:52










  • $begingroup$
    thanks it perfectly answers my query !
    $endgroup$
    – kira0705
    Jun 13 '18 at 12:57














9












9








9





$begingroup$

The operation $*$ is not associative, since:begin{align}bigl||2-1|-1bigr|&=0\&neq2\&=bigl|2-|1-1|bigr|.end{align}Therefore, $(G,*)$ is not a group.






share|cite|improve this answer











$endgroup$



The operation $*$ is not associative, since:begin{align}bigl||2-1|-1bigr|&=0\&neq2\&=bigl|2-|1-1|bigr|.end{align}Therefore, $(G,*)$ is not a group.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 13 '18 at 21:48

























answered Jun 13 '18 at 12:48









José Carlos SantosJosé Carlos Santos

157k22126227




157k22126227








  • 2




    $begingroup$
    Just wanted to add that associativity does work when $a=b=c$ or when any of the values is zero so to find a counter-example or prove associativity by brute force relatively few cases would need to be checked.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:52










  • $begingroup$
    thanks it perfectly answers my query !
    $endgroup$
    – kira0705
    Jun 13 '18 at 12:57














  • 2




    $begingroup$
    Just wanted to add that associativity does work when $a=b=c$ or when any of the values is zero so to find a counter-example or prove associativity by brute force relatively few cases would need to be checked.
    $endgroup$
    – CyclotomicField
    Jun 13 '18 at 12:52










  • $begingroup$
    thanks it perfectly answers my query !
    $endgroup$
    – kira0705
    Jun 13 '18 at 12:57








2




2




$begingroup$
Just wanted to add that associativity does work when $a=b=c$ or when any of the values is zero so to find a counter-example or prove associativity by brute force relatively few cases would need to be checked.
$endgroup$
– CyclotomicField
Jun 13 '18 at 12:52




$begingroup$
Just wanted to add that associativity does work when $a=b=c$ or when any of the values is zero so to find a counter-example or prove associativity by brute force relatively few cases would need to be checked.
$endgroup$
– CyclotomicField
Jun 13 '18 at 12:52












$begingroup$
thanks it perfectly answers my query !
$endgroup$
– kira0705
Jun 13 '18 at 12:57




$begingroup$
thanks it perfectly answers my query !
$endgroup$
– kira0705
Jun 13 '18 at 12:57


















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Ellipse (mathématiques)

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Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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