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

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

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 $a\circ e = a+k-k=a$

Verification $e\circ a = -k\circ a = -k+a+k=a$, as required.

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

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

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

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

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    $\begingroup$ Yes, it is group. (I solved this problem as homework in uni once) $\endgroup$ – Vladislav Mar 27 at 17:14
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Well, you have to show associativity as well.... $\endgroup$ – fleablood Mar 27 at 17:21
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    $\begingroup$ @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

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 $\phi(a+b)=\phi(a)\circ\phi(b)$ so that becomes a group isomorphism.

  • $\begingroup$ How can I learn which maps transfer a group to another? $\endgroup$ – mathreadler Mar 27 at 17:32
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    $\begingroup$ @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen. $\endgroup$ – rschwieb Mar 27 at 19:28

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 $\forall a \in \mathbb{Z}, a = (1 - k)^{\circ(a + k - 1)}$.

  • $\begingroup$ 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
  • $\begingroup$ "$\forall a \in \mathbb{Z} a \circ (1-k) = (a+1)$" 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 Mar 28 at 1:55

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.

  • $\begingroup$ 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

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