Styjun

What is the difference between "Grippe" and "Männergrippe"?

Filling a listlineplot with a texture

Can $! cause race conditions when used in scripts running in parallel?

Boot Windows from SAN

The Wires Underground

How were medieval castles built in swamps or marshes without draining them?

Why do proofs of Bernoulli's equation assume that forces on opposite ends point in different directions?

Prison offence - trespassing underwood fence

Where can/should I, as a high schooler, publish a paper regarding the derivation of a formula?

Ghidra: Prepend memory segment in assembly listing view

Does ostensible/specious make sense in this sentence?

Removal of て in Japanese novels

What is a natural problem in theory of computation?

"Opusculum hoc, quamdiu vixero, doctioribus emendandum offero."?

How to obtain a polynomial with these conditions?

Why did Khan ask Admiral James T. Kirk about Project Genesis?

Server Integrity Check CheckCommands question

"There were either twelve sexes or none."

Semantic difference between regular and irregular 'backen'

Limitations with dynamical systems vs. PDEs?

Add 2 new columns to existing dataframe using apply

How many birds in the bush?

When, exactly, does the Rogue Scout get to use their Skirmisher ability?

How to send bitcoin from an offline Bitcoin Core that doesn't have a synced balance?

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?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 groupDoes the set of all symmetries of a plane figure form a group under composition of functions?Let $G = mathbb R$ together with the operation $x*y = x + y + xy,; forall x,yin mathbb R$proving that $mathbbN$ is a group under operation $Δ$Real valued function concatenation as group operator? For which sets?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

8

2

$begingroup$

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 Mar 27 at 21:56

mathreadler

asked Mar 27 at 17:02

mathreadlermathreadler

16.4k88 gold badges2222 silver badges6363 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is group. (I solved this problem as homework in uni once)
  $endgroup$
  – Vladislav
  Mar 27 at 17:14
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How would we know if you've made a mistake when you haven't shared your work on the problem?
  $endgroup$
  – Shaun
  Mar 27 at 17:14
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 27 at 17:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you have to show associativity as well....
  $endgroup$
  – fleablood
  Mar 27 at 17:21
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 27 at 20:03
  

  
  
  
  

 | 
show 5 more comments

8

2

$begingroup$

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 Mar 27 at 21:56

mathreadler

asked Mar 27 at 17:02

mathreadlermathreadler

16.4k88 gold badges2222 silver badges6363 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is group. (I solved this problem as homework in uni once)
  $endgroup$
  – Vladislav
  Mar 27 at 17:14
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How would we know if you've made a mistake when you haven't shared your work on the problem?
  $endgroup$
  – Shaun
  Mar 27 at 17:14
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 27 at 17:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you have to show associativity as well....
  $endgroup$
  – fleablood
  Mar 27 at 17:21
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 27 at 20:03
  

  
  
  
  

 | 
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 Mar 27 at 21:56

mathreadler

asked Mar 27 at 17:02

mathreadlermathreadler

16.4k88 gold badges2222 silver badges6363 bronze badges

$endgroup$

share|cite|improve this question

edited Mar 27 at 21:56

mathreadler

asked Mar 27 at 17:02

mathreadlermathreadler

16.4k88 gold badges2222 silver badges6363 bronze badges

share|cite|improve this question

edited Mar 27 at 21:56

mathreadler

asked Mar 27 at 17:02

mathreadlermathreadler

16.4k88 gold badges2222 silver badges6363 bronze badges

asked Mar 27 at 17:02

mathreadlermathreadler

16.4k88 gold badges2222 silver badges6363 bronze badges

asked Mar 27 at 17:02

mathreadlermathreadler

16.4k88 gold badges2222 silver badges6363 bronze badges

3 Answers
3

active

oldest

votes

11

$begingroup$

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 Mar 27 at 17:27

rschwiebrschwieb

112k1212 gold badges111111 silver badges263263 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can I learn which maps transfer a group to another?
  $endgroup$
  – mathreadler
  Mar 27 at 17:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
  $endgroup$
  – rschwieb
  Mar 27 at 19:28
  
  

  
  
  
  

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

edited Mar 28 at 7:08

answered Mar 27 at 17:26

Yanior WegYanior Weg

3,79333 gold badges1919 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 27 at 22:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "$forall a in mathbbZ a circ (1-k) = (a+1)$" 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
  Mar 28 at 1:55
  

  
  
  
  

add a comment | 

3

$begingroup$

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 Mar 27 at 22:16

AcccumulationAcccumulation

7,96022 gold badges77 silver badges2020 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 27 at 22:28
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164781%2finteger-addition-constant-is-it-a-group%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

11

$begingroup$

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 Mar 27 at 17:27

rschwiebrschwieb

112k1212 gold badges111111 silver badges263263 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can I learn which maps transfer a group to another?
  $endgroup$
  – mathreadler
  Mar 27 at 17:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
  $endgroup$
  – rschwieb
  Mar 27 at 19:28
  
  

  
  
  
  

add a comment | 

11

$begingroup$

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 Mar 27 at 17:27

rschwiebrschwieb

112k1212 gold badges111111 silver badges263263 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can I learn which maps transfer a group to another?
  $endgroup$
  – mathreadler
  Mar 27 at 17:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
  $endgroup$
  – rschwieb
  Mar 27 at 19:28
  
  

  
  
  
  

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 Mar 27 at 17:27

rschwiebrschwieb

112k1212 gold badges111111 silver badges263263 bronze badges

$endgroup$

share|cite|improve this answer

answered Mar 27 at 17:27

rschwiebrschwieb

112k1212 gold badges111111 silver badges263263 bronze badges

answered Mar 27 at 17:27

rschwiebrschwieb

112k1212 gold badges111111 silver badges263263 bronze badges

answered Mar 27 at 17:27

rschwiebrschwieb

112k1212 gold badges111111 silver badges263263 bronze badges

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

edited Mar 28 at 7:08

answered Mar 27 at 17:26

Yanior WegYanior Weg

3,79333 gold badges1919 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 27 at 22:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "$forall a in mathbbZ a circ (1-k) = (a+1)$" 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
  Mar 28 at 1:55
  

  
  
  
  

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

edited Mar 28 at 7:08

answered Mar 27 at 17:26

Yanior WegYanior Weg

3,79333 gold badges1919 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 27 at 22:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "$forall a in mathbbZ a circ (1-k) = (a+1)$" 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
  Mar 28 at 1:55
  

  
  
  
  

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

edited Mar 28 at 7:08

answered Mar 27 at 17:26

Yanior WegYanior Weg

3,79333 gold badges1919 silver badges6666 bronze badges

$endgroup$

share|cite|improve this answer

edited Mar 28 at 7:08

answered Mar 27 at 17:26

Yanior WegYanior Weg

3,79333 gold badges1919 silver badges6666 bronze badges

answered Mar 27 at 17:26

Yanior WegYanior Weg

3,79333 gold badges1919 silver badges6666 bronze badges

answered Mar 27 at 17:26

Yanior WegYanior Weg

3,79333 gold badges1919 silver badges6666 bronze badges

3

$begingroup$

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 Mar 27 at 22:16

AcccumulationAcccumulation

7,96022 gold badges77 silver badges2020 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 27 at 22:28
  
  

  
  
  
  

add a comment | 

3

$begingroup$

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 Mar 27 at 22:16

AcccumulationAcccumulation

7,96022 gold badges77 silver badges2020 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 27 at 22:28
  
  

  
  
  
  

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 Mar 27 at 22:16

AcccumulationAcccumulation

7,96022 gold badges77 silver badges2020 bronze badges

$endgroup$

share|cite|improve this answer

answered Mar 27 at 22:16

AcccumulationAcccumulation

7,96022 gold badges77 silver badges2020 bronze badges

answered Mar 27 at 22:16

AcccumulationAcccumulation

7,96022 gold badges77 silver badges2020 bronze badges

answered Mar 27 at 22:16

AcccumulationAcccumulation

7,96022 gold badges77 silver badges2020 bronze badges