Check whether $G$ is group or not
$begingroup$
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
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
asked Jun 13 '18 at 12:41
kira0705kira0705
995
$endgroup$
|
show 6 more comments
$begingroup$
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
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
asked Jun 13 '18 at 12:41
kira0705kira0705
995
$endgroup$
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
|
show 6 more comments
$begingroup$
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
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
asked Jun 13 '18 at 12:41
kira0705kira0705
995
$endgroup$
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
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
asked Jun 13 '18 at 12:41
kira0705kira0705
995
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
asked Jun 13 '18 at 12:41
kira0705kira0705
995
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
edited Jan 7 at 16:27
José Carlos Santos
157k22126227
157k22126227
asked Jun 13 '18 at 12:41
kira0705kira0705
995
asked Jun 13 '18 at 12:41
kira0705kira0705
995
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
|
show 6 more comments
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
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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 $0ne2$.
answered Jun 13 '18 at 14:20
egregegreg
181k1485202
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jun 13 '18 at 12:48
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2818207%2fcheck-whether-g-is-group-or-not%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $0ne2$.
answered Jun 13 '18 at 14:20
egregegreg
181k1485202
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $0ne2$.
answered Jun 13 '18 at 14:20
egregegreg
181k1485202
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $0ne2$.
answered Jun 13 '18 at 14:20
egregegreg
181k1485202
$endgroup$
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 $0ne2$.
answered Jun 13 '18 at 14:20
egregegreg
181k1485202
answered Jun 13 '18 at 14:20
egregegreg
181k1485202
answered Jun 13 '18 at 14:20
egregegreg
181k1485202
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jun 13 '18 at 12:48
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jun 13 '18 at 12:48
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jun 13 '18 at 12:48
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
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.
answered Jun 13 '18 at 12:48
José Carlos SantosJosé Carlos Santos
157k22126227
edited Aug 13 '18 at 21:48
answered Jun 13 '18 at 12:48
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jun 13 '18 at 12:48
José Carlos SantosJosé Carlos Santos
157k22126227
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
add a comment |
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2818207%2fcheck-whether-g-is-group-or-not%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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