Integer addition + constant, is it a group?












7












$begingroup$


Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$



Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?



I have tried, and found that it fulfills all group axioms, but I might have made a mistake?



If it is a group, does it have a name?




My observations:




  • Closure is obvious as addition of integers is closed.


  • Identity If we take $e=-k$, then $acirc e = a+k-k=a$



Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.





  • Inverse would be $a^{-1} = -a-2k$, which is unique.


Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.





  • Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.


We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.






abstract-algebra group-theory arithmetic







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Yes, it is group. (I solved this problem as homework in uni once)
    $endgroup$
    – Vladislav
    8 hours ago








  • 1




    $begingroup$
    How would we know if you've made a mistake when you haven't shared your work on the problem?
    $endgroup$
    – Shaun
    8 hours ago






  • 1




    $begingroup$
    You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
    $endgroup$
    – fleablood
    8 hours ago






  • 1




    $begingroup$
    Well, you have to show associativity as well....
    $endgroup$
    – fleablood
    8 hours ago






  • 2




    $begingroup$
    @fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
    $endgroup$
    – TonyK
    6 hours ago
















7












$begingroup$


Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$



Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?



I have tried, and found that it fulfills all group axioms, but I might have made a mistake?



If it is a group, does it have a name?




My observations:




  • Closure is obvious as addition of integers is closed.


  • Identity If we take $e=-k$, then $acirc e = a+k-k=a$



Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.





  • Inverse would be $a^{-1} = -a-2k$, which is unique.


Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.





  • Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.


We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.






abstract-algebra group-theory arithmetic







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Yes, it is group. (I solved this problem as homework in uni once)
    $endgroup$
    – Vladislav
    8 hours ago








  • 1




    $begingroup$
    How would we know if you've made a mistake when you haven't shared your work on the problem?
    $endgroup$
    – Shaun
    8 hours ago






  • 1




    $begingroup$
    You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
    $endgroup$
    – fleablood
    8 hours ago






  • 1




    $begingroup$
    Well, you have to show associativity as well....
    $endgroup$
    – fleablood
    8 hours ago






  • 2




    $begingroup$
    @fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
    $endgroup$
    – TonyK
    6 hours ago














7












7








7


1



$begingroup$


Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$



Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?



I have tried, and found that it fulfills all group axioms, but I might have made a mistake?



If it is a group, does it have a name?




My observations:




  • Closure is obvious as addition of integers is closed.


  • Identity If we take $e=-k$, then $acirc e = a+k-k=a$



Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.





  • Inverse would be $a^{-1} = -a-2k$, which is unique.


Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.





  • Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.


We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.






abstract-algebra group-theory arithmetic







share|cite|improve this question











$endgroup$




Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$



Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?



I have tried, and found that it fulfills all group axioms, but I might have made a mistake?



If it is a group, does it have a name?




My observations:




  • Closure is obvious as addition of integers is closed.


  • Identity If we take $e=-k$, then $acirc e = a+k-k=a$



Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.





  • Inverse would be $a^{-1} = -a-2k$, which is unique.


Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.





  • Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.


We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.






abstract-algebra group-theory arithmetic




abstract-algebra group-theory arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago







mathreadler

















asked 9 hours ago









mathreadlermathreadler

15.3k72263




15.3k72263








  • 3




    $begingroup$
    Yes, it is group. (I solved this problem as homework in uni once)
    $endgroup$
    – Vladislav
    8 hours ago








  • 1




    $begingroup$
    How would we know if you've made a mistake when you haven't shared your work on the problem?
    $endgroup$
    – Shaun
    8 hours ago






  • 1




    $begingroup$
    You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
    $endgroup$
    – fleablood
    8 hours ago






  • 1




    $begingroup$
    Well, you have to show associativity as well....
    $endgroup$
    – fleablood
    8 hours ago






  • 2




    $begingroup$
    @fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
    $endgroup$
    – TonyK
    6 hours ago














  • 3




    $begingroup$
    Yes, it is group. (I solved this problem as homework in uni once)
    $endgroup$
    – Vladislav
    8 hours ago








  • 1




    $begingroup$
    How would we know if you've made a mistake when you haven't shared your work on the problem?
    $endgroup$
    – Shaun
    8 hours ago






  • 1




    $begingroup$
    You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
    $endgroup$
    – fleablood
    8 hours ago






  • 1




    $begingroup$
    Well, you have to show associativity as well....
    $endgroup$
    – fleablood
    8 hours ago






  • 2




    $begingroup$
    @fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
    $endgroup$
    – TonyK
    6 hours ago








3




3




$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
8 hours ago






$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
8 hours ago






1




1




$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
8 hours ago




$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
8 hours ago




1




1




$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
8 hours ago




$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
8 hours ago




1




1




$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
8 hours ago




$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
8 hours ago




2




2




$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
6 hours ago




$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
6 hours ago










3 Answers
3






active

oldest

votes


















8












$begingroup$

It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.



You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can I learn which maps transfer a group to another?
    $endgroup$
    – mathreadler
    8 hours ago






  • 1




    $begingroup$
    @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
    $endgroup$
    – rschwieb
    6 hours ago





















4












$begingroup$

Yes, your observations are correct - this is a group.



Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.



To prove that you can see, that $forall a in mathbb{Z} a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z} a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
    $endgroup$
    – mathreadler
    4 hours ago










  • $begingroup$
    "$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
    $endgroup$
    – Misha Lavrov
    9 mins ago



















2












$begingroup$

Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.



And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
    $endgroup$
    – mathreadler
    3 hours ago













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.



You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can I learn which maps transfer a group to another?
    $endgroup$
    – mathreadler
    8 hours ago






  • 1




    $begingroup$
    @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
    $endgroup$
    – rschwieb
    6 hours ago


















8












$begingroup$

It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.



You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can I learn which maps transfer a group to another?
    $endgroup$
    – mathreadler
    8 hours ago






  • 1




    $begingroup$
    @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
    $endgroup$
    – rschwieb
    6 hours ago
















8












8








8





$begingroup$

It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.



You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.






share|cite|improve this answer









$endgroup$



It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.



You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









rschwiebrschwieb

107k12103251




107k12103251












  • $begingroup$
    How can I learn which maps transfer a group to another?
    $endgroup$
    – mathreadler
    8 hours ago






  • 1




    $begingroup$
    @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
    $endgroup$
    – rschwieb
    6 hours ago




















  • $begingroup$
    How can I learn which maps transfer a group to another?
    $endgroup$
    – mathreadler
    8 hours ago






  • 1




    $begingroup$
    @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
    $endgroup$
    – rschwieb
    6 hours ago


















$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
8 hours ago




$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
8 hours ago




1




1




$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
6 hours ago






$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
6 hours ago













4












$begingroup$

Yes, your observations are correct - this is a group.



Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.



To prove that you can see, that $forall a in mathbb{Z} a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z} a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
    $endgroup$
    – mathreadler
    4 hours ago










  • $begingroup$
    "$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
    $endgroup$
    – Misha Lavrov
    9 mins ago
















4












$begingroup$

Yes, your observations are correct - this is a group.



Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.



To prove that you can see, that $forall a in mathbb{Z} a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z} a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
    $endgroup$
    – mathreadler
    4 hours ago










  • $begingroup$
    "$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
    $endgroup$
    – Misha Lavrov
    9 mins ago














4












4








4





$begingroup$

Yes, your observations are correct - this is a group.



Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.



To prove that you can see, that $forall a in mathbb{Z} a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z} a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer









$endgroup$



Yes, your observations are correct - this is a group.



Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.



To prove that you can see, that $forall a in mathbb{Z} a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z} a = (1 - k)^{circ(a + k - 1)}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









Yanior WegYanior Weg

2,77711346




2,77711346












  • $begingroup$
    Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
    $endgroup$
    – mathreadler
    4 hours ago










  • $begingroup$
    "$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
    $endgroup$
    – Misha Lavrov
    9 mins ago


















  • $begingroup$
    Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
    $endgroup$
    – mathreadler
    4 hours ago










  • $begingroup$
    "$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
    $endgroup$
    – Misha Lavrov
    9 mins ago
















$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
4 hours ago




$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
4 hours ago












$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
9 mins ago




$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
9 mins ago











2












$begingroup$

Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.



And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
    $endgroup$
    – mathreadler
    3 hours ago


















2












$begingroup$

Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.



And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
    $endgroup$
    – mathreadler
    3 hours ago
















2












2








2





$begingroup$

Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.



And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.






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$endgroup$



Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.



And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









AcccumulationAcccumulation

7,2052619




7,2052619












  • $begingroup$
    Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
    $endgroup$
    – mathreadler
    3 hours ago




















  • $begingroup$
    Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
    $endgroup$
    – mathreadler
    3 hours ago


















$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
3 hours ago






$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
3 hours ago




















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