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Integer addition + constant, is it a group?

A question about groups: may I substitute a binary operation with a function?Does $log$ of positive rationals form a group under addition?Equivalence class is a group - request for proof-verification.Proving that the unities of a ring form a group under multiplicationTo prove in a Group Left identity and left inverse implies right identity and right inverseAre $(mathbbR - 0, times)$ and $(mathbbR, +)$ the same group? What is its name?Show that $G$ is a groupProving translations in R and rotations in C are groups under compositionDoes the set of all symmetries of a plane figure form a group under composition of functions?Let $G = mathbb R$ together with the operation $x*y = x + y + xy,; forall x,yin mathbb R$

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

edited 4 hours ago

mathreadler

asked 9 hours ago

mathreadlermathreadler

15.3k72263

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is group. (I solved this problem as homework in uni once)
  $endgroup$
  – Vladislav
  9 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
  9 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
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you have to show associativity as well....
  $endgroup$
  – fleablood
  9 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
  

  
  
  
  

 | 
show 5 more comments

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

edited 4 hours ago

mathreadler

asked 9 hours ago

mathreadlermathreadler

15.3k72263

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is group. (I solved this problem as homework in uni once)
  $endgroup$
  – Vladislav
  9 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
  9 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
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you have to show associativity as well....
  $endgroup$
  – fleablood
  9 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
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

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

edited 4 hours ago

mathreadler

asked 9 hours ago

mathreadlermathreadler

15.3k72263

$endgroup$

share|cite|improve this question

edited 4 hours ago

mathreadler

asked 9 hours ago

mathreadlermathreadler

15.3k72263

share|cite|improve this question

edited 4 hours ago

mathreadler

asked 9 hours ago

mathreadlermathreadler

15.3k72263

asked 9 hours ago

mathreadlermathreadler

15.3k72263

asked 9 hours ago

mathreadlermathreadler

15.3k72263

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

answered 8 hours ago

rschwiebrschwieb

107k12103251

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

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 $forall a in mathbbZ a circ (1-k) = (a+1)$, which results in $forall a in mathbbZ a = (1 - k)^circ(a + k - 1)$.

share|cite|improve this answer

answered 9 hours ago

Yanior WegYanior Weg

2,77711346

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 4 hours ago

AcccumulationAcccumulation

7,2052619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

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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

answered 8 hours ago

rschwiebrschwieb

107k12103251

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

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

answered 8 hours ago

rschwiebrschwieb

107k12103251

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 8 hours ago

rschwiebrschwieb

107k12103251

$endgroup$

share|cite|improve this answer

answered 8 hours ago

rschwiebrschwieb

107k12103251

answered 8 hours ago

rschwiebrschwieb

107k12103251

answered 8 hours ago

rschwiebrschwieb

107k12103251

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 $forall a in mathbbZ a circ (1-k) = (a+1)$, which results in $forall a in mathbbZ a = (1 - k)^circ(a + k - 1)$.

share|cite|improve this answer

answered 9 hours ago

Yanior WegYanior Weg

2,77711346

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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 $forall a in mathbbZ a circ (1-k) = (a+1)$, which results in $forall a in mathbbZ a = (1 - k)^circ(a + k - 1)$.

share|cite|improve this answer

answered 9 hours ago

Yanior WegYanior Weg

2,77711346

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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 $forall a in mathbbZ a circ (1-k) = (a+1)$, which results in $forall a in mathbbZ a = (1 - k)^circ(a + k - 1)$.

share|cite|improve this answer

answered 9 hours ago

Yanior WegYanior Weg

2,77711346

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Yanior WegYanior Weg

2,77711346

answered 9 hours ago

Yanior WegYanior Weg

2,77711346

answered 9 hours ago

Yanior WegYanior Weg

2,77711346

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

answered 4 hours ago

AcccumulationAcccumulation

7,2052619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

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

answered 4 hours ago

AcccumulationAcccumulation

7,2052619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 4 hours ago

AcccumulationAcccumulation

7,2052619

$endgroup$

share|cite|improve this answer

answered 4 hours ago

AcccumulationAcccumulation

7,2052619

answered 4 hours ago

AcccumulationAcccumulation

7,2052619

answered 4 hours ago

AcccumulationAcccumulation

7,2052619