Mixing
How did Kazuo Matsuzaka come to think of this? (Group Theory)
$begingroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem about group theory in this book:
Let $G$ be a non-empty set.
"$/$" is a binary operation on $G$.
"$/$" satisfies the following 4 properties:
- $a/a = b/b.$
- $a/(b/b)=a.$
- $(a/a)/(b/c)=c/b.$
- $(a/c)/(b/c)=a/b.$
We define $a*b :=a/((b/b)/b)$.
Show that $(G, *)$ is a group.
I think this fact is very complicated.
How did Matsuzaka come to think of this?
How did Matsuzaka choose this 4 properties?
abstract-algebra group-theory
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
$endgroup$
|
show 1 more comment
$begingroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem about group theory in this book:
Let $G$ be a non-empty set.
"$/$" is a binary operation on $G$.
"$/$" satisfies the following 4 properties:
- $a/a = b/b.$
- $a/(b/b)=a.$
- $(a/a)/(b/c)=c/b.$
- $(a/c)/(b/c)=a/b.$
We define $a*b :=a/((b/b)/b)$.
Show that $(G, *)$ is a group.
I think this fact is very complicated.
How did Matsuzaka come to think of this?
How did Matsuzaka choose this 4 properties?
abstract-algebra group-theory
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
$endgroup$
2
$begingroup$
These are just the properties we would have if we already had a group and defined $a/b = ab^{-1}$.
$endgroup$
– Tobias Kildetoft
Jan 9 at 7:27
$begingroup$
Thank you very much, Tobias Kildetoft. But how did Matsuzaka choose this 4 properties?
$endgroup$
– tchappy ha
Jan 9 at 7:33
$begingroup$
If we define $a/b := ab^{-1}$, there are infinitely many properties that "$/$" satisfies.
$endgroup$
– tchappy ha
Jan 9 at 7:36
$begingroup$
This might be a better fit for hsm.se.
$endgroup$
– J.G.
Jan 9 at 7:39
3
$begingroup$
Surely this stuff goes back at least to G. Higman and B. H. Neumann in 1952, in their work which gives a single axiom for groups: $x / ((((x / x) / y) / z) / (((x / x) / x) / z)) = y$. (Higman and B. H. Neumann. Groups as groupoids with one law.)
$endgroup$
– ancientmathematician
Jan 9 at 8:03
|
show 1 more comment
$begingroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem about group theory in this book:
Let $G$ be a non-empty set.
"$/$" is a binary operation on $G$.
"$/$" satisfies the following 4 properties:
- $a/a = b/b.$
- $a/(b/b)=a.$
- $(a/a)/(b/c)=c/b.$
- $(a/c)/(b/c)=a/b.$
We define $a*b :=a/((b/b)/b)$.
Show that $(G, *)$ is a group.
I think this fact is very complicated.
How did Matsuzaka come to think of this?
How did Matsuzaka choose this 4 properties?
abstract-algebra group-theory
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
$endgroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem about group theory in this book:
Let $G$ be a non-empty set.
"$/$" is a binary operation on $G$.
"$/$" satisfies the following 4 properties:
- $a/a = b/b.$
- $a/(b/b)=a.$
- $(a/a)/(b/c)=c/b.$
- $(a/c)/(b/c)=a/b.$
We define $a*b :=a/((b/b)/b)$.
Show that $(G, *)$ is a group.
I think this fact is very complicated.
How did Matsuzaka come to think of this?
How did Matsuzaka choose this 4 properties?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
edited Jan 9 at 7:34
tchappy ha
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
asked Jan 9 at 7:21
tchappy hatchappy ha
647311
647311
2
$begingroup$
These are just the properties we would have if we already had a group and defined $a/b = ab^{-1}$.
$endgroup$
– Tobias Kildetoft
Jan 9 at 7:27
$begingroup$
Thank you very much, Tobias Kildetoft. But how did Matsuzaka choose this 4 properties?
$endgroup$
– tchappy ha
Jan 9 at 7:33
$begingroup$
If we define $a/b := ab^{-1}$, there are infinitely many properties that "$/$" satisfies.
$endgroup$
– tchappy ha
Jan 9 at 7:36
$begingroup$
This might be a better fit for hsm.se.
$endgroup$
– J.G.
Jan 9 at 7:39
3
$begingroup$
Surely this stuff goes back at least to G. Higman and B. H. Neumann in 1952, in their work which gives a single axiom for groups: $x / ((((x / x) / y) / z) / (((x / x) / x) / z)) = y$. (Higman and B. H. Neumann. Groups as groupoids with one law.)
$endgroup$
– ancientmathematician
Jan 9 at 8:03
|
show 1 more comment
2
$begingroup$
These are just the properties we would have if we already had a group and defined $a/b = ab^{-1}$.
$endgroup$
– Tobias Kildetoft
Jan 9 at 7:27
$begingroup$
Thank you very much, Tobias Kildetoft. But how did Matsuzaka choose this 4 properties?
$endgroup$
– tchappy ha
Jan 9 at 7:33
$begingroup$
If we define $a/b := ab^{-1}$, there are infinitely many properties that "$/$" satisfies.
$endgroup$
– tchappy ha
Jan 9 at 7:36
$begingroup$
This might be a better fit for hsm.se.
$endgroup$
– J.G.
Jan 9 at 7:39
3
$begingroup$
Surely this stuff goes back at least to G. Higman and B. H. Neumann in 1952, in their work which gives a single axiom for groups: $x / ((((x / x) / y) / z) / (((x / x) / x) / z)) = y$. (Higman and B. H. Neumann. Groups as groupoids with one law.)
$endgroup$
– ancientmathematician
Jan 9 at 8:03
2
2
$begingroup$
These are just the properties we would have if we already had a group and defined $a/b = ab^{-1}$.
$endgroup$
– Tobias Kildetoft
Jan 9 at 7:27
$begingroup$
These are just the properties we would have if we already had a group and defined $a/b = ab^{-1}$.
$endgroup$
– Tobias Kildetoft
Jan 9 at 7:27
$begingroup$
Thank you very much, Tobias Kildetoft. But how did Matsuzaka choose this 4 properties?
$endgroup$
– tchappy ha
Jan 9 at 7:33
$begingroup$
Thank you very much, Tobias Kildetoft. But how did Matsuzaka choose this 4 properties?
$endgroup$
– tchappy ha
Jan 9 at 7:33
$begingroup$
If we define $a/b := ab^{-1}$, there are infinitely many properties that "$/$" satisfies.
$endgroup$
– tchappy ha
Jan 9 at 7:36
$begingroup$
If we define $a/b := ab^{-1}$, there are infinitely many properties that "$/$" satisfies.
$endgroup$
– tchappy ha
Jan 9 at 7:36
$begingroup$
This might be a better fit for hsm.se.
$endgroup$
– J.G.
Jan 9 at 7:39
$begingroup$
This might be a better fit for hsm.se.
$endgroup$
– J.G.
Jan 9 at 7:39
3
3
$begingroup$
Surely this stuff goes back at least to G. Higman and B. H. Neumann in 1952, in their work which gives a single axiom for groups: $x / ((((x / x) / y) / z) / (((x / x) / x) / z)) = y$. (Higman and B. H. Neumann. Groups as groupoids with one law.)
$endgroup$
– ancientmathematician
Jan 9 at 8:03
$begingroup$
Surely this stuff goes back at least to G. Higman and B. H. Neumann in 1952, in their work which gives a single axiom for groups: $x / ((((x / x) / y) / z) / (((x / x) / x) / z)) = y$. (Higman and B. H. Neumann. Groups as groupoids with one law.)
$endgroup$
– ancientmathematician
Jan 9 at 8:03
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Let me try to guess what can have come into their mind.
First of all, as mentioned in the comments, these were found in order to mimick the properties of $(a,b) mapsto ab^{-1}$ in a group.
To get a group from the single operation $/$, we need to recover the neutral element. Bit $e= aa^{-1}=a/a$ for any $a$. So to get $e$ we need to have $a/a=b/b$.
If we call $e$ the common value of $b/b$, we need it to be a neutral element : $a/e = ae^{-1}= ae= a$. Hence $a/(b/b) = a$.
We need $e$ to also be neutral on the left : we need to say how $e/(b/c)$ relates to $b/c$, but since there is an inversion in the process, we need to put $c/b$.
At this point you have recovered the neutral element, and part of the inversion. Still you have no law concerning general $a,b,c$ that can account for associativity.
This is where 4 comes in. It relates a general $a,b,c$ and in terms of $ab^{-1}$ is : $(ac^{-1})(bc^{-1})^{-1} = ab^{-1}$. This means (thanks to 3. and to what it tells us about $(fg)^{-1}$ ) $(ac^{-1})(cb^{-1}) = ab^{-1}$. So this is a version of associativity in terms of $/$.
Now we have recovered the neutral element, the inversion, and associativity : we've got a group.
Note that analyzing where these laws come from tells you a great deal about how to prove that $(G, *)$ is a group, because you now have a clear route.
answered Jan 9 at 11:16
MaxMax
14.1k11142
$endgroup$
$begingroup$
Thank you very much, Max.
$endgroup$
– tchappy ha
Jan 10 at 12:23
add a comment |
$begingroup$
These are common properties of fractions.
$1 = 1$,
$a/1 = a$,
$1/(b/c) = c/b$, if $b neq 0, c neq 0$, and
$frac{a/c}{b/c} = a/b$, if $b neq 0, c neq 0$.
This is not all the properties of fractions of integers. So this describes an algebraic system with division of nonzero quantities only guaranteed to have the listed properties. If you think of another property of fractions that isn't a consequence of the above, then the algebraic system described above need not exhibit that property.
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
$endgroup$
$begingroup$
Thank you very much, Eric Towers.
$endgroup$
– tchappy ha
Jan 10 at 12:22
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let me try to guess what can have come into their mind.
First of all, as mentioned in the comments, these were found in order to mimick the properties of $(a,b) mapsto ab^{-1}$ in a group.
To get a group from the single operation $/$, we need to recover the neutral element. Bit $e= aa^{-1}=a/a$ for any $a$. So to get $e$ we need to have $a/a=b/b$.
If we call $e$ the common value of $b/b$, we need it to be a neutral element : $a/e = ae^{-1}= ae= a$. Hence $a/(b/b) = a$.
We need $e$ to also be neutral on the left : we need to say how $e/(b/c)$ relates to $b/c$, but since there is an inversion in the process, we need to put $c/b$.
At this point you have recovered the neutral element, and part of the inversion. Still you have no law concerning general $a,b,c$ that can account for associativity.
This is where 4 comes in. It relates a general $a,b,c$ and in terms of $ab^{-1}$ is : $(ac^{-1})(bc^{-1})^{-1} = ab^{-1}$. This means (thanks to 3. and to what it tells us about $(fg)^{-1}$ ) $(ac^{-1})(cb^{-1}) = ab^{-1}$. So this is a version of associativity in terms of $/$.
Now we have recovered the neutral element, the inversion, and associativity : we've got a group.
Note that analyzing where these laws come from tells you a great deal about how to prove that $(G, *)$ is a group, because you now have a clear route.
answered Jan 9 at 11:16
MaxMax
14.1k11142
$endgroup$
$begingroup$
Thank you very much, Max.
$endgroup$
– tchappy ha
Jan 10 at 12:23
add a comment |
$begingroup$
Let me try to guess what can have come into their mind.
First of all, as mentioned in the comments, these were found in order to mimick the properties of $(a,b) mapsto ab^{-1}$ in a group.
To get a group from the single operation $/$, we need to recover the neutral element. Bit $e= aa^{-1}=a/a$ for any $a$. So to get $e$ we need to have $a/a=b/b$.
If we call $e$ the common value of $b/b$, we need it to be a neutral element : $a/e = ae^{-1}= ae= a$. Hence $a/(b/b) = a$.
We need $e$ to also be neutral on the left : we need to say how $e/(b/c)$ relates to $b/c$, but since there is an inversion in the process, we need to put $c/b$.
At this point you have recovered the neutral element, and part of the inversion. Still you have no law concerning general $a,b,c$ that can account for associativity.
This is where 4 comes in. It relates a general $a,b,c$ and in terms of $ab^{-1}$ is : $(ac^{-1})(bc^{-1})^{-1} = ab^{-1}$. This means (thanks to 3. and to what it tells us about $(fg)^{-1}$ ) $(ac^{-1})(cb^{-1}) = ab^{-1}$. So this is a version of associativity in terms of $/$.
Now we have recovered the neutral element, the inversion, and associativity : we've got a group.
Note that analyzing where these laws come from tells you a great deal about how to prove that $(G, *)$ is a group, because you now have a clear route.
answered Jan 9 at 11:16
MaxMax
14.1k11142
$endgroup$
$begingroup$
Thank you very much, Max.
$endgroup$
– tchappy ha
Jan 10 at 12:23
add a comment |
$begingroup$
Let me try to guess what can have come into their mind.
First of all, as mentioned in the comments, these were found in order to mimick the properties of $(a,b) mapsto ab^{-1}$ in a group.
To get a group from the single operation $/$, we need to recover the neutral element. Bit $e= aa^{-1}=a/a$ for any $a$. So to get $e$ we need to have $a/a=b/b$.
If we call $e$ the common value of $b/b$, we need it to be a neutral element : $a/e = ae^{-1}= ae= a$. Hence $a/(b/b) = a$.
We need $e$ to also be neutral on the left : we need to say how $e/(b/c)$ relates to $b/c$, but since there is an inversion in the process, we need to put $c/b$.
At this point you have recovered the neutral element, and part of the inversion. Still you have no law concerning general $a,b,c$ that can account for associativity.
This is where 4 comes in. It relates a general $a,b,c$ and in terms of $ab^{-1}$ is : $(ac^{-1})(bc^{-1})^{-1} = ab^{-1}$. This means (thanks to 3. and to what it tells us about $(fg)^{-1}$ ) $(ac^{-1})(cb^{-1}) = ab^{-1}$. So this is a version of associativity in terms of $/$.
Now we have recovered the neutral element, the inversion, and associativity : we've got a group.
Note that analyzing where these laws come from tells you a great deal about how to prove that $(G, *)$ is a group, because you now have a clear route.
answered Jan 9 at 11:16
MaxMax
14.1k11142
$endgroup$
Let me try to guess what can have come into their mind.
First of all, as mentioned in the comments, these were found in order to mimick the properties of $(a,b) mapsto ab^{-1}$ in a group.
To get a group from the single operation $/$, we need to recover the neutral element. Bit $e= aa^{-1}=a/a$ for any $a$. So to get $e$ we need to have $a/a=b/b$.
If we call $e$ the common value of $b/b$, we need it to be a neutral element : $a/e = ae^{-1}= ae= a$. Hence $a/(b/b) = a$.
We need $e$ to also be neutral on the left : we need to say how $e/(b/c)$ relates to $b/c$, but since there is an inversion in the process, we need to put $c/b$.
At this point you have recovered the neutral element, and part of the inversion. Still you have no law concerning general $a,b,c$ that can account for associativity.
This is where 4 comes in. It relates a general $a,b,c$ and in terms of $ab^{-1}$ is : $(ac^{-1})(bc^{-1})^{-1} = ab^{-1}$. This means (thanks to 3. and to what it tells us about $(fg)^{-1}$ ) $(ac^{-1})(cb^{-1}) = ab^{-1}$. So this is a version of associativity in terms of $/$.
Now we have recovered the neutral element, the inversion, and associativity : we've got a group.
Note that analyzing where these laws come from tells you a great deal about how to prove that $(G, *)$ is a group, because you now have a clear route.
answered Jan 9 at 11:16
MaxMax
14.1k11142
answered Jan 9 at 11:16
MaxMax
14.1k11142
answered Jan 9 at 11:16
MaxMax
14.1k11142
answered Jan 9 at 11:16
MaxMax
14.1k11142
14.1k11142
$begingroup$
Thank you very much, Max.
$endgroup$
– tchappy ha
Jan 10 at 12:23
add a comment |
$begingroup$
Thank you very much, Max.
$endgroup$
– tchappy ha
Jan 10 at 12:23
$begingroup$
Thank you very much, Max.
$endgroup$
– tchappy ha
Jan 10 at 12:23
$begingroup$
Thank you very much, Max.
$endgroup$
– tchappy ha
Jan 10 at 12:23
add a comment |
$begingroup$
These are common properties of fractions.
$1 = 1$,
$a/1 = a$,
$1/(b/c) = c/b$, if $b neq 0, c neq 0$, and
$frac{a/c}{b/c} = a/b$, if $b neq 0, c neq 0$.
This is not all the properties of fractions of integers. So this describes an algebraic system with division of nonzero quantities only guaranteed to have the listed properties. If you think of another property of fractions that isn't a consequence of the above, then the algebraic system described above need not exhibit that property.
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
$endgroup$
$begingroup$
Thank you very much, Eric Towers.
$endgroup$
– tchappy ha
Jan 10 at 12:22
add a comment |
$begingroup$
These are common properties of fractions.
$1 = 1$,
$a/1 = a$,
$1/(b/c) = c/b$, if $b neq 0, c neq 0$, and
$frac{a/c}{b/c} = a/b$, if $b neq 0, c neq 0$.
This is not all the properties of fractions of integers. So this describes an algebraic system with division of nonzero quantities only guaranteed to have the listed properties. If you think of another property of fractions that isn't a consequence of the above, then the algebraic system described above need not exhibit that property.
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
$endgroup$
$begingroup$
Thank you very much, Eric Towers.
$endgroup$
– tchappy ha
Jan 10 at 12:22
add a comment |
$begingroup$
These are common properties of fractions.
$1 = 1$,
$a/1 = a$,
$1/(b/c) = c/b$, if $b neq 0, c neq 0$, and
$frac{a/c}{b/c} = a/b$, if $b neq 0, c neq 0$.
This is not all the properties of fractions of integers. So this describes an algebraic system with division of nonzero quantities only guaranteed to have the listed properties. If you think of another property of fractions that isn't a consequence of the above, then the algebraic system described above need not exhibit that property.
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
$endgroup$
These are common properties of fractions.
$1 = 1$,
$a/1 = a$,
$1/(b/c) = c/b$, if $b neq 0, c neq 0$, and
$frac{a/c}{b/c} = a/b$, if $b neq 0, c neq 0$.
This is not all the properties of fractions of integers. So this describes an algebraic system with division of nonzero quantities only guaranteed to have the listed properties. If you think of another property of fractions that isn't a consequence of the above, then the algebraic system described above need not exhibit that property.
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
answered Jan 9 at 7:42
Eric TowersEric Towers
32.5k22370
32.5k22370
$begingroup$
Thank you very much, Eric Towers.
$endgroup$
– tchappy ha
Jan 10 at 12:22
add a comment |
$begingroup$
Thank you very much, Eric Towers.
$endgroup$
– tchappy ha
Jan 10 at 12:22
$begingroup$
Thank you very much, Eric Towers.
$endgroup$
– tchappy ha
Jan 10 at 12:22
$begingroup$
Thank you very much, Eric Towers.
$endgroup$
– tchappy ha
Jan 10 at 12:22
add a comment |
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$begingroup$
These are just the properties we would have if we already had a group and defined $a/b = ab^{-1}$.
$endgroup$
– Tobias Kildetoft
Jan 9 at 7:27
$begingroup$
Thank you very much, Tobias Kildetoft. But how did Matsuzaka choose this 4 properties?
$endgroup$
– tchappy ha
Jan 9 at 7:33
$begingroup$
If we define $a/b := ab^{-1}$, there are infinitely many properties that "$/$" satisfies.
$endgroup$
– tchappy ha
Jan 9 at 7:36
$begingroup$
This might be a better fit for hsm.se.
$endgroup$
– J.G.
Jan 9 at 7:39
3
$begingroup$
Surely this stuff goes back at least to G. Higman and B. H. Neumann in 1952, in their work which gives a single axiom for groups: $x / ((((x / x) / y) / z) / (((x / x) / x) / z)) = y$. (Higman and B. H. Neumann. Groups as groupoids with one law.)
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– ancientmathematician
Jan 9 at 8:03