Mixing
Need to prove that (S,*) defined by the binary operation a*b = a+b+ab is an abelian group on S = R {1}
$begingroup$
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.
The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
group-theory abelian-groups
asked Oct 3 '13 at 4:20
MikMik
1721217
$endgroup$
add a comment |
$begingroup$
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.
The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
group-theory abelian-groups
asked Oct 3 '13 at 4:20
MikMik
1721217
$endgroup$
$begingroup$
"Because if 0 is the identity element, then this group won't have inverses." What about $Bbb Z$ with $+$ and $0$?
$endgroup$
– Pedro Tamaroff♦
Oct 3 '13 at 4:22
$begingroup$
By $R$, do you mean $mathbb R$?
$endgroup$
– Ian Coley
Oct 3 '13 at 4:22
$begingroup$
Similar question: math.stackexchange.com/questions/980196/…
$endgroup$
– Martin Sleziak
Oct 23 '14 at 8:59
add a comment |
$begingroup$
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.
The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
group-theory abelian-groups
asked Oct 3 '13 at 4:20
MikMik
1721217
$endgroup$
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.
The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
group-theory abelian-groups
group-theory abelian-groups
asked Oct 3 '13 at 4:20
MikMik
1721217
asked Oct 3 '13 at 4:20
MikMik
1721217
asked Oct 3 '13 at 4:20
MikMik
1721217
asked Oct 3 '13 at 4:20
MikMik
1721217
asked Oct 3 '13 at 4:20
MikMik
1721217
1721217
$begingroup$
"Because if 0 is the identity element, then this group won't have inverses." What about $Bbb Z$ with $+$ and $0$?
$endgroup$
– Pedro Tamaroff♦
Oct 3 '13 at 4:22
$begingroup$
By $R$, do you mean $mathbb R$?
$endgroup$
– Ian Coley
Oct 3 '13 at 4:22
$begingroup$
Similar question: math.stackexchange.com/questions/980196/…
$endgroup$
– Martin Sleziak
Oct 23 '14 at 8:59
add a comment |
$begingroup$
"Because if 0 is the identity element, then this group won't have inverses." What about $Bbb Z$ with $+$ and $0$?
$endgroup$
– Pedro Tamaroff♦
Oct 3 '13 at 4:22
$begingroup$
By $R$, do you mean $mathbb R$?
$endgroup$
– Ian Coley
Oct 3 '13 at 4:22
$begingroup$
Similar question: math.stackexchange.com/questions/980196/…
$endgroup$
– Martin Sleziak
Oct 23 '14 at 8:59
$begingroup$
"Because if 0 is the identity element, then this group won't have inverses." What about $Bbb Z$ with $+$ and $0$?
$endgroup$
– Pedro Tamaroff♦
Oct 3 '13 at 4:22
$begingroup$
"Because if 0 is the identity element, then this group won't have inverses." What about $Bbb Z$ with $+$ and $0$?
$endgroup$
– Pedro Tamaroff♦
Oct 3 '13 at 4:22
$begingroup$
By $R$, do you mean $mathbb R$?
$endgroup$
– Ian Coley
Oct 3 '13 at 4:22
$begingroup$
By $R$, do you mean $mathbb R$?
$endgroup$
– Ian Coley
Oct 3 '13 at 4:22
$begingroup$
Similar question: math.stackexchange.com/questions/980196/…
$endgroup$
– Martin Sleziak
Oct 23 '14 at 8:59
$begingroup$
Similar question: math.stackexchange.com/questions/980196/…
$endgroup$
– Martin Sleziak
Oct 23 '14 at 8:59
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I assume you mean $S=mathbb Rsetminus {-1}$.
Take $f:Sto mathbb R^*$ given by $f(x)=x+1$. This is a bijection.
Now note that for $a,bin S$, we have $a*b= a+b+ab=(a+1)(b+1)-1=f^{-1}(f(a)f(b))$.
What $f$ does is to rename the elements of $S$ as elements of $mathbb R^*$ and operate there. So $S$ is a group because it has been forced to be isomorphic to $mathbb R^*$ via $f$. That's all there is to it.
For instance, $0in S$ is the identity because $f(0)=1$ and $1$ is the identity of $mathbb R^*$.
This is an example of a pullback. See https://math.stackexchange.com/a/373743/589
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
$endgroup$
1
$begingroup$
+1, but that's not all there is to it! It is the series expansion around the origin of the formal multiplicative group.
$endgroup$
– Bruno Joyal
Oct 3 '13 at 4:33
$begingroup$
@BrunoJoyal, interesting!
$endgroup$
– goblin
Feb 9 '16 at 12:33
add a comment |
$begingroup$
I believe you meant to write $S=mathbb{R}backslash{-1}$
$0$ is indeed the identity element since for any $ain S$, $a * 0=a+0+a.0=a$
For $b$ to be the inverse of $a$, we require $a * b=0$.
Hence $a+b+a.b=0$
$b+a.b=-a$
$b(1+a)=-a$
$b=frac{-a}{1+a}$
which is fine, since $a$ can't be $-1$ (since it's not an element of $S$).
answered Oct 3 '13 at 4:29
johnjohn
4,910923
$endgroup$
add a comment |
$begingroup$
In general for any associative ring $R$, the circle operation $x circ y=x+y+xy$ defines a monoid structure on the underlying set of $R$, and the invertible elements in this monoid form a group called the adjoint group of the ring $R$, let us denote it by $Q(R)$.
If $R$ has an identity for the multiplication, $Q(R)$ is isomorphic to the group of units of $R$ under the mapping $x mapsto 1+x$, equivalently the set ${,y-1 in R mid y mbox{ is a unit in }R,}$ forms a group under our circle operation.
If we are working with a field $mathbb R$, we have only to exclude $0-1=-1$ from this set. In particular, the elements of $mathbb{R}setminus{-1}$ form a group under our circle operation.
edited Jan 14 at 16:41
Orat
2,80021231
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
$endgroup$
add a comment |
$begingroup$
I think the identity has to be 0: $a*e = a+e+ae=a Rightarrow e+ae=0 Rightarrow e(1+a)=0 Rightarrow e=0 $.
The inverse of $a$ is: $a*a^{-1}=0 Rightarrow a+a^{-1}+aa^{-1}=0 Rightarrow a^{-1} = -frac{a}{1+a}$.
Since $a neq -1$, so for all $a$, $a^{-1}$ always exists.
answered Oct 3 '13 at 4:26
user60177user60177
1715
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume you mean $S=mathbb Rsetminus {-1}$.
Take $f:Sto mathbb R^*$ given by $f(x)=x+1$. This is a bijection.
Now note that for $a,bin S$, we have $a*b= a+b+ab=(a+1)(b+1)-1=f^{-1}(f(a)f(b))$.
What $f$ does is to rename the elements of $S$ as elements of $mathbb R^*$ and operate there. So $S$ is a group because it has been forced to be isomorphic to $mathbb R^*$ via $f$. That's all there is to it.
For instance, $0in S$ is the identity because $f(0)=1$ and $1$ is the identity of $mathbb R^*$.
This is an example of a pullback. See https://math.stackexchange.com/a/373743/589
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
$endgroup$
1
$begingroup$
+1, but that's not all there is to it! It is the series expansion around the origin of the formal multiplicative group.
$endgroup$
– Bruno Joyal
Oct 3 '13 at 4:33
$begingroup$
@BrunoJoyal, interesting!
$endgroup$
– goblin
Feb 9 '16 at 12:33
add a comment |
$begingroup$
I assume you mean $S=mathbb Rsetminus {-1}$.
Take $f:Sto mathbb R^*$ given by $f(x)=x+1$. This is a bijection.
Now note that for $a,bin S$, we have $a*b= a+b+ab=(a+1)(b+1)-1=f^{-1}(f(a)f(b))$.
What $f$ does is to rename the elements of $S$ as elements of $mathbb R^*$ and operate there. So $S$ is a group because it has been forced to be isomorphic to $mathbb R^*$ via $f$. That's all there is to it.
For instance, $0in S$ is the identity because $f(0)=1$ and $1$ is the identity of $mathbb R^*$.
This is an example of a pullback. See https://math.stackexchange.com/a/373743/589
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
$endgroup$
1
$begingroup$
+1, but that's not all there is to it! It is the series expansion around the origin of the formal multiplicative group.
$endgroup$
– Bruno Joyal
Oct 3 '13 at 4:33
$begingroup$
@BrunoJoyal, interesting!
$endgroup$
– goblin
Feb 9 '16 at 12:33
add a comment |
$begingroup$
I assume you mean $S=mathbb Rsetminus {-1}$.
Take $f:Sto mathbb R^*$ given by $f(x)=x+1$. This is a bijection.
Now note that for $a,bin S$, we have $a*b= a+b+ab=(a+1)(b+1)-1=f^{-1}(f(a)f(b))$.
What $f$ does is to rename the elements of $S$ as elements of $mathbb R^*$ and operate there. So $S$ is a group because it has been forced to be isomorphic to $mathbb R^*$ via $f$. That's all there is to it.
For instance, $0in S$ is the identity because $f(0)=1$ and $1$ is the identity of $mathbb R^*$.
This is an example of a pullback. See https://math.stackexchange.com/a/373743/589
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
$endgroup$
I assume you mean $S=mathbb Rsetminus {-1}$.
Take $f:Sto mathbb R^*$ given by $f(x)=x+1$. This is a bijection.
Now note that for $a,bin S$, we have $a*b= a+b+ab=(a+1)(b+1)-1=f^{-1}(f(a)f(b))$.
What $f$ does is to rename the elements of $S$ as elements of $mathbb R^*$ and operate there. So $S$ is a group because it has been forced to be isomorphic to $mathbb R^*$ via $f$. That's all there is to it.
For instance, $0in S$ is the identity because $f(0)=1$ and $1$ is the identity of $mathbb R^*$.
This is an example of a pullback. See https://math.stackexchange.com/a/373743/589
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
answered Oct 3 '13 at 4:31
lhflhf
165k10171396
165k10171396
1
$begingroup$
+1, but that's not all there is to it! It is the series expansion around the origin of the formal multiplicative group.
$endgroup$
– Bruno Joyal
Oct 3 '13 at 4:33
$begingroup$
@BrunoJoyal, interesting!
$endgroup$
– goblin
Feb 9 '16 at 12:33
add a comment |
1
$begingroup$
+1, but that's not all there is to it! It is the series expansion around the origin of the formal multiplicative group.
$endgroup$
– Bruno Joyal
Oct 3 '13 at 4:33
$begingroup$
@BrunoJoyal, interesting!
$endgroup$
– goblin
Feb 9 '16 at 12:33
1
1
$begingroup$
+1, but that's not all there is to it! It is the series expansion around the origin of the formal multiplicative group.
$endgroup$
– Bruno Joyal
Oct 3 '13 at 4:33
$begingroup$
+1, but that's not all there is to it! It is the series expansion around the origin of the formal multiplicative group.
$endgroup$
– Bruno Joyal
Oct 3 '13 at 4:33
$begingroup$
@BrunoJoyal, interesting!
$endgroup$
– goblin
Feb 9 '16 at 12:33
$begingroup$
@BrunoJoyal, interesting!
$endgroup$
– goblin
Feb 9 '16 at 12:33
add a comment |
$begingroup$
I believe you meant to write $S=mathbb{R}backslash{-1}$
$0$ is indeed the identity element since for any $ain S$, $a * 0=a+0+a.0=a$
For $b$ to be the inverse of $a$, we require $a * b=0$.
Hence $a+b+a.b=0$
$b+a.b=-a$
$b(1+a)=-a$
$b=frac{-a}{1+a}$
which is fine, since $a$ can't be $-1$ (since it's not an element of $S$).
answered Oct 3 '13 at 4:29
johnjohn
4,910923
$endgroup$
add a comment |
$begingroup$
I believe you meant to write $S=mathbb{R}backslash{-1}$
$0$ is indeed the identity element since for any $ain S$, $a * 0=a+0+a.0=a$
For $b$ to be the inverse of $a$, we require $a * b=0$.
Hence $a+b+a.b=0$
$b+a.b=-a$
$b(1+a)=-a$
$b=frac{-a}{1+a}$
which is fine, since $a$ can't be $-1$ (since it's not an element of $S$).
answered Oct 3 '13 at 4:29
johnjohn
4,910923
$endgroup$
add a comment |
$begingroup$
I believe you meant to write $S=mathbb{R}backslash{-1}$
$0$ is indeed the identity element since for any $ain S$, $a * 0=a+0+a.0=a$
For $b$ to be the inverse of $a$, we require $a * b=0$.
Hence $a+b+a.b=0$
$b+a.b=-a$
$b(1+a)=-a$
$b=frac{-a}{1+a}$
which is fine, since $a$ can't be $-1$ (since it's not an element of $S$).
answered Oct 3 '13 at 4:29
johnjohn
4,910923
$endgroup$
I believe you meant to write $S=mathbb{R}backslash{-1}$
$0$ is indeed the identity element since for any $ain S$, $a * 0=a+0+a.0=a$
For $b$ to be the inverse of $a$, we require $a * b=0$.
Hence $a+b+a.b=0$
$b+a.b=-a$
$b(1+a)=-a$
$b=frac{-a}{1+a}$
which is fine, since $a$ can't be $-1$ (since it's not an element of $S$).
answered Oct 3 '13 at 4:29
johnjohn
4,910923
answered Oct 3 '13 at 4:29
johnjohn
4,910923
answered Oct 3 '13 at 4:29
johnjohn
4,910923
answered Oct 3 '13 at 4:29
johnjohn
4,910923
4,910923
add a comment |
add a comment |
$begingroup$
In general for any associative ring $R$, the circle operation $x circ y=x+y+xy$ defines a monoid structure on the underlying set of $R$, and the invertible elements in this monoid form a group called the adjoint group of the ring $R$, let us denote it by $Q(R)$.
If $R$ has an identity for the multiplication, $Q(R)$ is isomorphic to the group of units of $R$ under the mapping $x mapsto 1+x$, equivalently the set ${,y-1 in R mid y mbox{ is a unit in }R,}$ forms a group under our circle operation.
If we are working with a field $mathbb R$, we have only to exclude $0-1=-1$ from this set. In particular, the elements of $mathbb{R}setminus{-1}$ form a group under our circle operation.
edited Jan 14 at 16:41
Orat
2,80021231
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
$endgroup$
add a comment |
$begingroup$
In general for any associative ring $R$, the circle operation $x circ y=x+y+xy$ defines a monoid structure on the underlying set of $R$, and the invertible elements in this monoid form a group called the adjoint group of the ring $R$, let us denote it by $Q(R)$.
If $R$ has an identity for the multiplication, $Q(R)$ is isomorphic to the group of units of $R$ under the mapping $x mapsto 1+x$, equivalently the set ${,y-1 in R mid y mbox{ is a unit in }R,}$ forms a group under our circle operation.
If we are working with a field $mathbb R$, we have only to exclude $0-1=-1$ from this set. In particular, the elements of $mathbb{R}setminus{-1}$ form a group under our circle operation.
edited Jan 14 at 16:41
Orat
2,80021231
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
$endgroup$
add a comment |
$begingroup$
In general for any associative ring $R$, the circle operation $x circ y=x+y+xy$ defines a monoid structure on the underlying set of $R$, and the invertible elements in this monoid form a group called the adjoint group of the ring $R$, let us denote it by $Q(R)$.
If $R$ has an identity for the multiplication, $Q(R)$ is isomorphic to the group of units of $R$ under the mapping $x mapsto 1+x$, equivalently the set ${,y-1 in R mid y mbox{ is a unit in }R,}$ forms a group under our circle operation.
If we are working with a field $mathbb R$, we have only to exclude $0-1=-1$ from this set. In particular, the elements of $mathbb{R}setminus{-1}$ form a group under our circle operation.
edited Jan 14 at 16:41
Orat
2,80021231
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
$endgroup$
In general for any associative ring $R$, the circle operation $x circ y=x+y+xy$ defines a monoid structure on the underlying set of $R$, and the invertible elements in this monoid form a group called the adjoint group of the ring $R$, let us denote it by $Q(R)$.
If $R$ has an identity for the multiplication, $Q(R)$ is isomorphic to the group of units of $R$ under the mapping $x mapsto 1+x$, equivalently the set ${,y-1 in R mid y mbox{ is a unit in }R,}$ forms a group under our circle operation.
If we are working with a field $mathbb R$, we have only to exclude $0-1=-1$ from this set. In particular, the elements of $mathbb{R}setminus{-1}$ form a group under our circle operation.
edited Jan 14 at 16:41
Orat
2,80021231
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
edited Jan 14 at 16:41
Orat
2,80021231
edited Jan 14 at 16:41
Orat
2,80021231
edited Jan 14 at 16:41
Orat
2,80021231
2,80021231
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
answered Oct 3 '13 at 9:09
Yassine GuerboussaYassine Guerboussa
1,219711
1,219711
add a comment |
add a comment |
$begingroup$
I think the identity has to be 0: $a*e = a+e+ae=a Rightarrow e+ae=0 Rightarrow e(1+a)=0 Rightarrow e=0 $.
The inverse of $a$ is: $a*a^{-1}=0 Rightarrow a+a^{-1}+aa^{-1}=0 Rightarrow a^{-1} = -frac{a}{1+a}$.
Since $a neq -1$, so for all $a$, $a^{-1}$ always exists.
answered Oct 3 '13 at 4:26
user60177user60177
1715
$endgroup$
add a comment |
$begingroup$
I think the identity has to be 0: $a*e = a+e+ae=a Rightarrow e+ae=0 Rightarrow e(1+a)=0 Rightarrow e=0 $.
The inverse of $a$ is: $a*a^{-1}=0 Rightarrow a+a^{-1}+aa^{-1}=0 Rightarrow a^{-1} = -frac{a}{1+a}$.
Since $a neq -1$, so for all $a$, $a^{-1}$ always exists.
answered Oct 3 '13 at 4:26
user60177user60177
1715
$endgroup$
add a comment |
$begingroup$
I think the identity has to be 0: $a*e = a+e+ae=a Rightarrow e+ae=0 Rightarrow e(1+a)=0 Rightarrow e=0 $.
The inverse of $a$ is: $a*a^{-1}=0 Rightarrow a+a^{-1}+aa^{-1}=0 Rightarrow a^{-1} = -frac{a}{1+a}$.
Since $a neq -1$, so for all $a$, $a^{-1}$ always exists.
answered Oct 3 '13 at 4:26
user60177user60177
1715
$endgroup$
I think the identity has to be 0: $a*e = a+e+ae=a Rightarrow e+ae=0 Rightarrow e(1+a)=0 Rightarrow e=0 $.
The inverse of $a$ is: $a*a^{-1}=0 Rightarrow a+a^{-1}+aa^{-1}=0 Rightarrow a^{-1} = -frac{a}{1+a}$.
Since $a neq -1$, so for all $a$, $a^{-1}$ always exists.
answered Oct 3 '13 at 4:26
user60177user60177
1715
answered Oct 3 '13 at 4:26
user60177user60177
1715
answered Oct 3 '13 at 4:26
user60177user60177
1715
answered Oct 3 '13 at 4:26
user60177user60177
1715
1715
add a comment |
add a comment |
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$begingroup$
"Because if 0 is the identity element, then this group won't have inverses." What about $Bbb Z$ with $+$ and $0$?
$endgroup$
– Pedro Tamaroff♦
Oct 3 '13 at 4:22
$begingroup$
By $R$, do you mean $mathbb R$?
$endgroup$
– Ian Coley
Oct 3 '13 at 4:22
$begingroup$
Similar question: math.stackexchange.com/questions/980196/…
$endgroup$
– Martin Sleziak
Oct 23 '14 at 8:59