Mixing
If $G$ is a group and $a,bin G$ and $e$ neutral element why $ab = aeb$?
$begingroup$
The Problem I have if this would be true then one could also say $aaa^{-1}b=ab$.
Why do I know that there don't exist counterexamples for which the inequality holds?
group-theory
asked Feb 1 at 18:10
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
The Problem I have if this would be true then one could also say $aaa^{-1}b=ab$.
Why do I know that there don't exist counterexamples for which the inequality holds?
group-theory
asked Feb 1 at 18:10
RM777RM777
38312
$endgroup$
1
$begingroup$
$aeb = a(eb) = ab$, and $aaa^{-1}b = a(aa^{-1})b = aeb = ab$.
$endgroup$
– Nicolas
Feb 1 at 18:15
add a comment |
$begingroup$
The Problem I have if this would be true then one could also say $aaa^{-1}b=ab$.
Why do I know that there don't exist counterexamples for which the inequality holds?
group-theory
asked Feb 1 at 18:10
RM777RM777
38312
$endgroup$
The Problem I have if this would be true then one could also say $aaa^{-1}b=ab$.
Why do I know that there don't exist counterexamples for which the inequality holds?
group-theory
group-theory
asked Feb 1 at 18:10
RM777RM777
38312
asked Feb 1 at 18:10
RM777RM777
38312
asked Feb 1 at 18:10
RM777RM777
38312
asked Feb 1 at 18:10
RM777RM777
38312
asked Feb 1 at 18:10
RM777RM777
38312
38312
1
$begingroup$
$aeb = a(eb) = ab$, and $aaa^{-1}b = a(aa^{-1})b = aeb = ab$.
$endgroup$
– Nicolas
Feb 1 at 18:15
add a comment |
1
$begingroup$
$aeb = a(eb) = ab$, and $aaa^{-1}b = a(aa^{-1})b = aeb = ab$.
$endgroup$
– Nicolas
Feb 1 at 18:15
1
1
$begingroup$
$aeb = a(eb) = ab$, and $aaa^{-1}b = a(aa^{-1})b = aeb = ab$.
$endgroup$
– Nicolas
Feb 1 at 18:15
$begingroup$
$aeb = a(eb) = ab$, and $aaa^{-1}b = a(aa^{-1})b = aeb = ab$.
$endgroup$
– Nicolas
Feb 1 at 18:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Among the group axioms you also have associativity. That allowed you to write
$$aeb = a(eb) = ab $$
I assume by neutral you mean the identity. The inverse element axiom states that every element has an inverse with the property
$$xx^{-1} = x^{-1}x = e $$
Thus we can use associativity as follows
$$aaa^{-1}b = a(aa^{-1})b = aeb = a(eb) = ab$$
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
$endgroup$
$begingroup$
The part of this proof:math.stackexchange.com/questions/310351/… where the first answer says and simillary we get $yM=n'M$ makes use of this. I was not sure if the proof requires this Transformation and whether the Transformation was correct
$endgroup$
– RM777
Feb 1 at 18:14
add a comment |
$begingroup$
If $e$ is neutral element then $$forall x. ex =x $$
so $eb$ transforms into $b$ (because $eb = b$)
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
$endgroup$
add a comment |
$begingroup$
$aaa^{-1}b = a(aa^{-1})b = aeb = (ae)b = ab$
So yes, you CAN say that, and you'd be one hundred percent correct when you do say that.
Why do I know that there don't exist counterexamples for which the inequality holds?
The same way you know that there aren't any counter examples to $ktimes frac 1k =1$ when $k ne 0$.
BY DEFINITION $aa^{-1} = e$ and BY DEFINITION $aeb = (ae)b = ab$. So that is a PROOF.
answered Feb 1 at 18:23
fleabloodfleablood
1
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Among the group axioms you also have associativity. That allowed you to write
$$aeb = a(eb) = ab $$
I assume by neutral you mean the identity. The inverse element axiom states that every element has an inverse with the property
$$xx^{-1} = x^{-1}x = e $$
Thus we can use associativity as follows
$$aaa^{-1}b = a(aa^{-1})b = aeb = a(eb) = ab$$
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
$endgroup$
$begingroup$
The part of this proof:math.stackexchange.com/questions/310351/… where the first answer says and simillary we get $yM=n'M$ makes use of this. I was not sure if the proof requires this Transformation and whether the Transformation was correct
$endgroup$
– RM777
Feb 1 at 18:14
add a comment |
$begingroup$
Among the group axioms you also have associativity. That allowed you to write
$$aeb = a(eb) = ab $$
I assume by neutral you mean the identity. The inverse element axiom states that every element has an inverse with the property
$$xx^{-1} = x^{-1}x = e $$
Thus we can use associativity as follows
$$aaa^{-1}b = a(aa^{-1})b = aeb = a(eb) = ab$$
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
$endgroup$
$begingroup$
The part of this proof:math.stackexchange.com/questions/310351/… where the first answer says and simillary we get $yM=n'M$ makes use of this. I was not sure if the proof requires this Transformation and whether the Transformation was correct
$endgroup$
– RM777
Feb 1 at 18:14
add a comment |
$begingroup$
Among the group axioms you also have associativity. That allowed you to write
$$aeb = a(eb) = ab $$
I assume by neutral you mean the identity. The inverse element axiom states that every element has an inverse with the property
$$xx^{-1} = x^{-1}x = e $$
Thus we can use associativity as follows
$$aaa^{-1}b = a(aa^{-1})b = aeb = a(eb) = ab$$
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
$endgroup$
Among the group axioms you also have associativity. That allowed you to write
$$aeb = a(eb) = ab $$
I assume by neutral you mean the identity. The inverse element axiom states that every element has an inverse with the property
$$xx^{-1} = x^{-1}x = e $$
Thus we can use associativity as follows
$$aaa^{-1}b = a(aa^{-1})b = aeb = a(eb) = ab$$
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
edited Feb 1 at 18:14
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
answered Feb 1 at 18:12
Alvin LepikAlvin Lepik
2,85611024
2,85611024
$begingroup$
The part of this proof:math.stackexchange.com/questions/310351/… where the first answer says and simillary we get $yM=n'M$ makes use of this. I was not sure if the proof requires this Transformation and whether the Transformation was correct
$endgroup$
– RM777
Feb 1 at 18:14
add a comment |
$begingroup$
The part of this proof:math.stackexchange.com/questions/310351/… where the first answer says and simillary we get $yM=n'M$ makes use of this. I was not sure if the proof requires this Transformation and whether the Transformation was correct
$endgroup$
– RM777
Feb 1 at 18:14
$begingroup$
The part of this proof:math.stackexchange.com/questions/310351/… where the first answer says and simillary we get $yM=n'M$ makes use of this. I was not sure if the proof requires this Transformation and whether the Transformation was correct
$endgroup$
– RM777
Feb 1 at 18:14
$begingroup$
The part of this proof:math.stackexchange.com/questions/310351/… where the first answer says and simillary we get $yM=n'M$ makes use of this. I was not sure if the proof requires this Transformation and whether the Transformation was correct
$endgroup$
– RM777
Feb 1 at 18:14
add a comment |
$begingroup$
If $e$ is neutral element then $$forall x. ex =x $$
so $eb$ transforms into $b$ (because $eb = b$)
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
$endgroup$
add a comment |
$begingroup$
If $e$ is neutral element then $$forall x. ex =x $$
so $eb$ transforms into $b$ (because $eb = b$)
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
$endgroup$
add a comment |
$begingroup$
If $e$ is neutral element then $$forall x. ex =x $$
so $eb$ transforms into $b$ (because $eb = b$)
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
$endgroup$
If $e$ is neutral element then $$forall x. ex =x $$
so $eb$ transforms into $b$ (because $eb = b$)
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
answered Feb 1 at 18:16
VirtualUserVirtualUser
1,321317
1,321317
add a comment |
add a comment |
$begingroup$
$aaa^{-1}b = a(aa^{-1})b = aeb = (ae)b = ab$
So yes, you CAN say that, and you'd be one hundred percent correct when you do say that.
Why do I know that there don't exist counterexamples for which the inequality holds?
The same way you know that there aren't any counter examples to $ktimes frac 1k =1$ when $k ne 0$.
BY DEFINITION $aa^{-1} = e$ and BY DEFINITION $aeb = (ae)b = ab$. So that is a PROOF.
answered Feb 1 at 18:23
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
$aaa^{-1}b = a(aa^{-1})b = aeb = (ae)b = ab$
So yes, you CAN say that, and you'd be one hundred percent correct when you do say that.
Why do I know that there don't exist counterexamples for which the inequality holds?
The same way you know that there aren't any counter examples to $ktimes frac 1k =1$ when $k ne 0$.
BY DEFINITION $aa^{-1} = e$ and BY DEFINITION $aeb = (ae)b = ab$. So that is a PROOF.
answered Feb 1 at 18:23
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
$aaa^{-1}b = a(aa^{-1})b = aeb = (ae)b = ab$
So yes, you CAN say that, and you'd be one hundred percent correct when you do say that.
Why do I know that there don't exist counterexamples for which the inequality holds?
The same way you know that there aren't any counter examples to $ktimes frac 1k =1$ when $k ne 0$.
BY DEFINITION $aa^{-1} = e$ and BY DEFINITION $aeb = (ae)b = ab$. So that is a PROOF.
answered Feb 1 at 18:23
fleabloodfleablood
1
$endgroup$
$aaa^{-1}b = a(aa^{-1})b = aeb = (ae)b = ab$
So yes, you CAN say that, and you'd be one hundred percent correct when you do say that.
Why do I know that there don't exist counterexamples for which the inequality holds?
The same way you know that there aren't any counter examples to $ktimes frac 1k =1$ when $k ne 0$.
BY DEFINITION $aa^{-1} = e$ and BY DEFINITION $aeb = (ae)b = ab$. So that is a PROOF.
answered Feb 1 at 18:23
fleabloodfleablood
1
answered Feb 1 at 18:23
fleabloodfleablood
1
answered Feb 1 at 18:23
fleabloodfleablood
1
answered Feb 1 at 18:23
fleabloodfleablood
1
1
add a comment |
add a comment |
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1
$begingroup$
$aeb = a(eb) = ab$, and $aaa^{-1}b = a(aa^{-1})b = aeb = ab$.
$endgroup$
– Nicolas
Feb 1 at 18:15