Build abelian group containing a set $K$ under an associative, commutative operation $*$ with an identity but...
We are given a set K of elements and an operation * . For every element in the set, there exists an inverse element (not necessarily in the set). There are three (additional) rules:
1) K contains e (neutral element) such that for every x: x * e = e * x = x
2) K is associative
3) K is commutative
The task is to build an abelian group G such that G contains K
OR to prove that it is impossible.
The task seems to be intuitively simple, but the concrete proof seems to be unobvious.
group-theory finite-groups abelian-groups semigroups magma
edited Nov 28 at 23:00
Shaun
8,344113578
asked Dec 14 '16 at 12:42
mathjunkie
1047
add a comment |
We are given a set K of elements and an operation * . For every element in the set, there exists an inverse element (not necessarily in the set). There are three (additional) rules:
1) K contains e (neutral element) such that for every x: x * e = e * x = x
2) K is associative
3) K is commutative
The task is to build an abelian group G such that G contains K
OR to prove that it is impossible.
The task seems to be intuitively simple, but the concrete proof seems to be unobvious.
group-theory finite-groups abelian-groups semigroups magma
edited Nov 28 at 23:00
Shaun
8,344113578
asked Dec 14 '16 at 12:42
mathjunkie
1047
I, be submitted an edit. Yes, e is "truly" neutral, and an inverse element exists for every element.
– mathjunkie
Dec 14 '16 at 12:49
Why didn't you list the existence of inverses among the rules in the first place?
– Arthur
Dec 14 '16 at 12:50
Fixed my mistake.
– mathjunkie
Dec 14 '16 at 12:52
add a comment |
We are given a set K of elements and an operation * . For every element in the set, there exists an inverse element (not necessarily in the set). There are three (additional) rules:
1) K contains e (neutral element) such that for every x: x * e = e * x = x
2) K is associative
3) K is commutative
The task is to build an abelian group G such that G contains K
OR to prove that it is impossible.
The task seems to be intuitively simple, but the concrete proof seems to be unobvious.
group-theory finite-groups abelian-groups semigroups magma
edited Nov 28 at 23:00
Shaun
8,344113578
asked Dec 14 '16 at 12:42
mathjunkie
1047
We are given a set K of elements and an operation * . For every element in the set, there exists an inverse element (not necessarily in the set). There are three (additional) rules:
1) K contains e (neutral element) such that for every x: x * e = e * x = x
2) K is associative
3) K is commutative
The task is to build an abelian group G such that G contains K
OR to prove that it is impossible.
The task seems to be intuitively simple, but the concrete proof seems to be unobvious.
group-theory finite-groups abelian-groups semigroups magma
group-theory finite-groups abelian-groups semigroups magma
edited Nov 28 at 23:00
Shaun
8,344113578
asked Dec 14 '16 at 12:42
mathjunkie
1047
edited Nov 28 at 23:00
Shaun
8,344113578
asked Dec 14 '16 at 12:42
mathjunkie
1047
edited Nov 28 at 23:00
Shaun
8,344113578
edited Nov 28 at 23:00
Shaun
8,344113578
edited Nov 28 at 23:00
Shaun
8,344113578
8,344113578
asked Dec 14 '16 at 12:42
mathjunkie
1047
asked Dec 14 '16 at 12:42
mathjunkie
1047
asked Dec 14 '16 at 12:42
mathjunkie
1047
1047
I, be submitted an edit. Yes, e is "truly" neutral, and an inverse element exists for every element.
– mathjunkie
Dec 14 '16 at 12:49
Why didn't you list the existence of inverses among the rules in the first place?
– Arthur
Dec 14 '16 at 12:50
Fixed my mistake.
– mathjunkie
Dec 14 '16 at 12:52
add a comment |
I, be submitted an edit. Yes, e is "truly" neutral, and an inverse element exists for every element.
– mathjunkie
Dec 14 '16 at 12:49
Why didn't you list the existence of inverses among the rules in the first place?
– Arthur
Dec 14 '16 at 12:50
Fixed my mistake.
– mathjunkie
Dec 14 '16 at 12:52
I, be submitted an edit. Yes, e is "truly" neutral, and an inverse element exists for every element.
– mathjunkie
Dec 14 '16 at 12:49
I, be submitted an edit. Yes, e is "truly" neutral, and an inverse element exists for every element.
– mathjunkie
Dec 14 '16 at 12:49
Why didn't you list the existence of inverses among the rules in the first place?
– Arthur
Dec 14 '16 at 12:50
Why didn't you list the existence of inverses among the rules in the first place?
– Arthur
Dec 14 '16 at 12:50
Fixed my mistake.
– mathjunkie
Dec 14 '16 at 12:52
Fixed my mistake.
– mathjunkie
Dec 14 '16 at 12:52
add a comment |
1 Answer
1
active
oldest
votes
It's not necessarily possible, since you have no guarantee of the existence of inverses. Specifically, if $K$ is the set of integers and * is multiplication, then it satisfies all three rules. Alternatively, if $K = {0,1,2,3,ldots}$ and * is addition, you also have all three rules satisfied.
What you have with those three rules is called a (commutative) monoid. Remove condition 1) and you have a semigroup.
Edit: After adding the rule that each element has an inverse, your (now four) rules are exactly the rules of an abelian group.
answered Dec 14 '16 at 12:45
Arthur
110k7105186
But (a) the inverse is not necessarily in the set and (b) x1*x2 might not lie in the set
– mathjunkie
Dec 14 '16 at 12:57
OK, so we're not even assuming that the operation is closed. Well, then, there is really nothing more you can do unless you have a concrete example in front of you.
– Arthur
Dec 14 '16 at 13:00
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
It's not necessarily possible, since you have no guarantee of the existence of inverses. Specifically, if $K$ is the set of integers and * is multiplication, then it satisfies all three rules. Alternatively, if $K = {0,1,2,3,ldots}$ and * is addition, you also have all three rules satisfied.
What you have with those three rules is called a (commutative) monoid. Remove condition 1) and you have a semigroup.
Edit: After adding the rule that each element has an inverse, your (now four) rules are exactly the rules of an abelian group.
answered Dec 14 '16 at 12:45
Arthur
110k7105186
But (a) the inverse is not necessarily in the set and (b) x1*x2 might not lie in the set
– mathjunkie
Dec 14 '16 at 12:57
OK, so we're not even assuming that the operation is closed. Well, then, there is really nothing more you can do unless you have a concrete example in front of you.
– Arthur
Dec 14 '16 at 13:00
add a comment |
It's not necessarily possible, since you have no guarantee of the existence of inverses. Specifically, if $K$ is the set of integers and * is multiplication, then it satisfies all three rules. Alternatively, if $K = {0,1,2,3,ldots}$ and * is addition, you also have all three rules satisfied.
What you have with those three rules is called a (commutative) monoid. Remove condition 1) and you have a semigroup.
Edit: After adding the rule that each element has an inverse, your (now four) rules are exactly the rules of an abelian group.
answered Dec 14 '16 at 12:45
Arthur
110k7105186
But (a) the inverse is not necessarily in the set and (b) x1*x2 might not lie in the set
– mathjunkie
Dec 14 '16 at 12:57
OK, so we're not even assuming that the operation is closed. Well, then, there is really nothing more you can do unless you have a concrete example in front of you.
– Arthur
Dec 14 '16 at 13:00
add a comment |
It's not necessarily possible, since you have no guarantee of the existence of inverses. Specifically, if $K$ is the set of integers and * is multiplication, then it satisfies all three rules. Alternatively, if $K = {0,1,2,3,ldots}$ and * is addition, you also have all three rules satisfied.
What you have with those three rules is called a (commutative) monoid. Remove condition 1) and you have a semigroup.
Edit: After adding the rule that each element has an inverse, your (now four) rules are exactly the rules of an abelian group.
answered Dec 14 '16 at 12:45
Arthur
110k7105186
It's not necessarily possible, since you have no guarantee of the existence of inverses. Specifically, if $K$ is the set of integers and * is multiplication, then it satisfies all three rules. Alternatively, if $K = {0,1,2,3,ldots}$ and * is addition, you also have all three rules satisfied.
What you have with those three rules is called a (commutative) monoid. Remove condition 1) and you have a semigroup.
Edit: After adding the rule that each element has an inverse, your (now four) rules are exactly the rules of an abelian group.
answered Dec 14 '16 at 12:45
Arthur
110k7105186
edited Dec 14 '16 at 12:51
answered Dec 14 '16 at 12:45
Arthur
110k7105186
answered Dec 14 '16 at 12:45
Arthur
110k7105186
answered Dec 14 '16 at 12:45
Arthur
110k7105186
110k7105186
But (a) the inverse is not necessarily in the set and (b) x1*x2 might not lie in the set
– mathjunkie
Dec 14 '16 at 12:57
OK, so we're not even assuming that the operation is closed. Well, then, there is really nothing more you can do unless you have a concrete example in front of you.
– Arthur
Dec 14 '16 at 13:00
add a comment |
But (a) the inverse is not necessarily in the set and (b) x1*x2 might not lie in the set
– mathjunkie
Dec 14 '16 at 12:57
OK, so we're not even assuming that the operation is closed. Well, then, there is really nothing more you can do unless you have a concrete example in front of you.
– Arthur
Dec 14 '16 at 13:00
But (a) the inverse is not necessarily in the set and (b) x1*x2 might not lie in the set
– mathjunkie
Dec 14 '16 at 12:57
But (a) the inverse is not necessarily in the set and (b) x1*x2 might not lie in the set
– mathjunkie
Dec 14 '16 at 12:57
OK, so we're not even assuming that the operation is closed. Well, then, there is really nothing more you can do unless you have a concrete example in front of you.
– Arthur
Dec 14 '16 at 13:00
OK, so we're not even assuming that the operation is closed. Well, then, there is really nothing more you can do unless you have a concrete example in front of you.
– Arthur
Dec 14 '16 at 13:00
add a comment |
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I, be submitted an edit. Yes, e is "truly" neutral, and an inverse element exists for every element.
– mathjunkie
Dec 14 '16 at 12:49
Why didn't you list the existence of inverses among the rules in the first place?
– Arthur
Dec 14 '16 at 12:50
Fixed my mistake.
– mathjunkie
Dec 14 '16 at 12:52