Integer addition + constant, is it a group?

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.