Group theory: How does binary operation define its associated set?
$begingroup$
I am trying to do a little reading in group theory. An exercise in the book I am reading asks for a proof of:
If $(G,circ)$ and $H(,circ)$ are groups both associated with associative operation $circ$, $G = H$. And, therefore, the set of a group is defined by the given binary operation.
I haven't been able to find a proof. I thought I might find a reductio ad absurdum by assuming that $G$ had at least one member not in $H$ (or vice versa) and derive some contradiction from that. Haven't had any luck. Anyone know how to prove this?
group-theory
edited Jan 7 at 15:25
A.Γ.
22.6k32656
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
$endgroup$
add a comment |
$begingroup$
I am trying to do a little reading in group theory. An exercise in the book I am reading asks for a proof of:
If $(G,circ)$ and $H(,circ)$ are groups both associated with associative operation $circ$, $G = H$. And, therefore, the set of a group is defined by the given binary operation.
I haven't been able to find a proof. I thought I might find a reductio ad absurdum by assuming that $G$ had at least one member not in $H$ (or vice versa) and derive some contradiction from that. Haven't had any luck. Anyone know how to prove this?
group-theory
edited Jan 7 at 15:25
A.Γ.
22.6k32656
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
$endgroup$
$begingroup$
Subgroups are definitely a thing, and they are usually conidered to have the same operation. So this means we have to ask: What is the exact, formal definition your book has of "binary operation"? (And if it starts with "a function", then we need to ask what its definition of "function" is too.)
$endgroup$
– Arthur
Jan 7 at 15:23
add a comment |
$begingroup$
I am trying to do a little reading in group theory. An exercise in the book I am reading asks for a proof of:
If $(G,circ)$ and $H(,circ)$ are groups both associated with associative operation $circ$, $G = H$. And, therefore, the set of a group is defined by the given binary operation.
I haven't been able to find a proof. I thought I might find a reductio ad absurdum by assuming that $G$ had at least one member not in $H$ (or vice versa) and derive some contradiction from that. Haven't had any luck. Anyone know how to prove this?
group-theory
edited Jan 7 at 15:25
A.Γ.
22.6k32656
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
$endgroup$
I am trying to do a little reading in group theory. An exercise in the book I am reading asks for a proof of:
If $(G,circ)$ and $H(,circ)$ are groups both associated with associative operation $circ$, $G = H$. And, therefore, the set of a group is defined by the given binary operation.
I haven't been able to find a proof. I thought I might find a reductio ad absurdum by assuming that $G$ had at least one member not in $H$ (or vice versa) and derive some contradiction from that. Haven't had any luck. Anyone know how to prove this?
group-theory
group-theory
edited Jan 7 at 15:25
A.Γ.
22.6k32656
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
edited Jan 7 at 15:25
A.Γ.
22.6k32656
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
edited Jan 7 at 15:25
A.Γ.
22.6k32656
edited Jan 7 at 15:25
A.Γ.
22.6k32656
edited Jan 7 at 15:25
A.Γ.
22.6k32656
22.6k32656
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
asked Jan 7 at 15:21
bob.sacamentobob.sacamento
2,4261819
2,4261819
$begingroup$
Subgroups are definitely a thing, and they are usually conidered to have the same operation. So this means we have to ask: What is the exact, formal definition your book has of "binary operation"? (And if it starts with "a function", then we need to ask what its definition of "function" is too.)
$endgroup$
– Arthur
Jan 7 at 15:23
add a comment |
$begingroup$
Subgroups are definitely a thing, and they are usually conidered to have the same operation. So this means we have to ask: What is the exact, formal definition your book has of "binary operation"? (And if it starts with "a function", then we need to ask what its definition of "function" is too.)
$endgroup$
– Arthur
Jan 7 at 15:23
$begingroup$
Subgroups are definitely a thing, and they are usually conidered to have the same operation. So this means we have to ask: What is the exact, formal definition your book has of "binary operation"? (And if it starts with "a function", then we need to ask what its definition of "function" is too.)
$endgroup$
– Arthur
Jan 7 at 15:23
$begingroup$
Subgroups are definitely a thing, and they are usually conidered to have the same operation. So this means we have to ask: What is the exact, formal definition your book has of "binary operation"? (And if it starts with "a function", then we need to ask what its definition of "function" is too.)
$endgroup$
– Arthur
Jan 7 at 15:23
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
My first instinct was to say that this does not make sense because the same "binary operation" can be used for many groups, e.g. $+$ for $mathbb{Z}subseteq mathbb{Q}subseteq mathbb{R}subseteq mathbb{C}$. However, the definition of a binary operation on a set $X$ is that it is a map $Xtimes Xto X$. A function is specified by the data of its domain, codomain, and "rule." So if two groups $G$ and $H$ have the same binary operation, their codomains align: $G=H$ as do the domains: $Gtimes G=Htimes H$. In particular, $(G,circ)=(H,circ)$ as groups.
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
$begingroup$
I don't understand what is there to prove here. Isn't it trivial? If it is the same set and same operation then it is obviously the same group. (of course same operation implies same identity and same inverses as well)
$endgroup$
– Mark
Jan 7 at 15:33
1
$begingroup$
Yes I agree, if this exercise makes sense at all, then it is trivial. I can only speculate that the point of the exercise is to ensure that the student understands the definition of an operation.
$endgroup$
– Derek Holt
Jan 7 at 15:37
$begingroup$
I agree, also. With this interpretation the exercise is much too "nitpicky." Without this interpretation it is senseless.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 15:55
add a comment |
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$begingroup$
My first instinct was to say that this does not make sense because the same "binary operation" can be used for many groups, e.g. $+$ for $mathbb{Z}subseteq mathbb{Q}subseteq mathbb{R}subseteq mathbb{C}$. However, the definition of a binary operation on a set $X$ is that it is a map $Xtimes Xto X$. A function is specified by the data of its domain, codomain, and "rule." So if two groups $G$ and $H$ have the same binary operation, their codomains align: $G=H$ as do the domains: $Gtimes G=Htimes H$. In particular, $(G,circ)=(H,circ)$ as groups.
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
$begingroup$
I don't understand what is there to prove here. Isn't it trivial? If it is the same set and same operation then it is obviously the same group. (of course same operation implies same identity and same inverses as well)
$endgroup$
– Mark
Jan 7 at 15:33
1
$begingroup$
Yes I agree, if this exercise makes sense at all, then it is trivial. I can only speculate that the point of the exercise is to ensure that the student understands the definition of an operation.
$endgroup$
– Derek Holt
Jan 7 at 15:37
$begingroup$
I agree, also. With this interpretation the exercise is much too "nitpicky." Without this interpretation it is senseless.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 15:55
add a comment |
$begingroup$
My first instinct was to say that this does not make sense because the same "binary operation" can be used for many groups, e.g. $+$ for $mathbb{Z}subseteq mathbb{Q}subseteq mathbb{R}subseteq mathbb{C}$. However, the definition of a binary operation on a set $X$ is that it is a map $Xtimes Xto X$. A function is specified by the data of its domain, codomain, and "rule." So if two groups $G$ and $H$ have the same binary operation, their codomains align: $G=H$ as do the domains: $Gtimes G=Htimes H$. In particular, $(G,circ)=(H,circ)$ as groups.
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
$begingroup$
I don't understand what is there to prove here. Isn't it trivial? If it is the same set and same operation then it is obviously the same group. (of course same operation implies same identity and same inverses as well)
$endgroup$
– Mark
Jan 7 at 15:33
1
$begingroup$
Yes I agree, if this exercise makes sense at all, then it is trivial. I can only speculate that the point of the exercise is to ensure that the student understands the definition of an operation.
$endgroup$
– Derek Holt
Jan 7 at 15:37
$begingroup$
I agree, also. With this interpretation the exercise is much too "nitpicky." Without this interpretation it is senseless.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 15:55
add a comment |
$begingroup$
My first instinct was to say that this does not make sense because the same "binary operation" can be used for many groups, e.g. $+$ for $mathbb{Z}subseteq mathbb{Q}subseteq mathbb{R}subseteq mathbb{C}$. However, the definition of a binary operation on a set $X$ is that it is a map $Xtimes Xto X$. A function is specified by the data of its domain, codomain, and "rule." So if two groups $G$ and $H$ have the same binary operation, their codomains align: $G=H$ as do the domains: $Gtimes G=Htimes H$. In particular, $(G,circ)=(H,circ)$ as groups.
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
My first instinct was to say that this does not make sense because the same "binary operation" can be used for many groups, e.g. $+$ for $mathbb{Z}subseteq mathbb{Q}subseteq mathbb{R}subseteq mathbb{C}$. However, the definition of a binary operation on a set $X$ is that it is a map $Xtimes Xto X$. A function is specified by the data of its domain, codomain, and "rule." So if two groups $G$ and $H$ have the same binary operation, their codomains align: $G=H$ as do the domains: $Gtimes G=Htimes H$. In particular, $(G,circ)=(H,circ)$ as groups.
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 7 at 15:28
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
9,71741640
$begingroup$
I don't understand what is there to prove here. Isn't it trivial? If it is the same set and same operation then it is obviously the same group. (of course same operation implies same identity and same inverses as well)
$endgroup$
– Mark
Jan 7 at 15:33
1
$begingroup$
Yes I agree, if this exercise makes sense at all, then it is trivial. I can only speculate that the point of the exercise is to ensure that the student understands the definition of an operation.
$endgroup$
– Derek Holt
Jan 7 at 15:37
$begingroup$
I agree, also. With this interpretation the exercise is much too "nitpicky." Without this interpretation it is senseless.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 15:55
add a comment |
$begingroup$
I don't understand what is there to prove here. Isn't it trivial? If it is the same set and same operation then it is obviously the same group. (of course same operation implies same identity and same inverses as well)
$endgroup$
– Mark
Jan 7 at 15:33
1
$begingroup$
Yes I agree, if this exercise makes sense at all, then it is trivial. I can only speculate that the point of the exercise is to ensure that the student understands the definition of an operation.
$endgroup$
– Derek Holt
Jan 7 at 15:37
$begingroup$
I agree, also. With this interpretation the exercise is much too "nitpicky." Without this interpretation it is senseless.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 15:55
$begingroup$
I don't understand what is there to prove here. Isn't it trivial? If it is the same set and same operation then it is obviously the same group. (of course same operation implies same identity and same inverses as well)
$endgroup$
– Mark
Jan 7 at 15:33
$begingroup$
I don't understand what is there to prove here. Isn't it trivial? If it is the same set and same operation then it is obviously the same group. (of course same operation implies same identity and same inverses as well)
$endgroup$
– Mark
Jan 7 at 15:33
1
1
$begingroup$
Yes I agree, if this exercise makes sense at all, then it is trivial. I can only speculate that the point of the exercise is to ensure that the student understands the definition of an operation.
$endgroup$
– Derek Holt
Jan 7 at 15:37
$begingroup$
Yes I agree, if this exercise makes sense at all, then it is trivial. I can only speculate that the point of the exercise is to ensure that the student understands the definition of an operation.
$endgroup$
– Derek Holt
Jan 7 at 15:37
$begingroup$
I agree, also. With this interpretation the exercise is much too "nitpicky." Without this interpretation it is senseless.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 15:55
$begingroup$
I agree, also. With this interpretation the exercise is much too "nitpicky." Without this interpretation it is senseless.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 15:55
add a comment |
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$begingroup$
Subgroups are definitely a thing, and they are usually conidered to have the same operation. So this means we have to ask: What is the exact, formal definition your book has of "binary operation"? (And if it starts with "a function", then we need to ask what its definition of "function" is too.)
$endgroup$
– Arthur
Jan 7 at 15:23