Prove that $\Bbb Z\times\Bbb Z/\langle (1,1)\rangle$ is isomorphic to $\Bbb Z$,

Question

Prove that $\Bbb Z\times\Bbb Z/\langle (1,1)\rangle$ is isomorphic to $\Bbb Z$.

First, I define the function $ f: \Bbb Z \times \Bbb Z \to \Bbb Z $, given by $ f ((a, b)) = a-b $. So the kernel of that function is $ \langle (1,1) \rangle $. I want to use irst Group Isomorphism Theorem, but I would need to prove that $ f $ is a homomorphism. But the only property I am missing is that of $ f (g^{-1}) = (f (g))^{-1} $. Someone help me with that part and could you tell me if my reasoning is correct.

Answer 1

For a group homomorphism it is only necessary to show that the operations commute with the mapping.

Then it follows automatically that the neutral element is transported and inverses are mapped as well as explicated above.

So in your case, you need only to show that $f((a,b) + (c,d)) = f((a,b)) + f((c,d))$.

Answer 2

Let's clarify a couple of things:

Technically, since a group (from the point of view of general/universal algebra) is a set with three operations (the multiplication/composition/addition, the inverse, and the identity element), a homomorphism of groups is required to respect all three operations. That is, within the general framework of universal algebra, you would ask that a morphism of groups, $f\colon (G,\cdot,{}^{-1},e)\to (H,\odot,{}^{\ominus1},1)$ is a function of underlying sets $f\colon G\to H$ such that for all $x,y\in G$,

1. $f(x\cdot y) = f(x)\odot f(y)$;
2. $f(x^{-1}) = (f(x))^{\ominus1}$; and
3. $f(e) = 1$.

Many books define homomorphism this way, to fit within the general framework.

However, there is a theorem that says that in fact you only need 1; that is, any semigroup homomorphism between groups is in fact a group homomorphism:

Theorem Let $(G,\cdot,{}^{-1},e)$ and $(H,\odot,{}^{\ominus1},1)$ be groups. A function $f\colon G\to H$ is a group homomorphism if and only if for every $x,y\in G$, $f(x\cdot y) = f(x)\odot f(y)$.

Some books define "group homomorphism" this way, requiring only that it be a semigroup homomorphism. In practice, once you've established this theorem, that is the way to go: no need to prove that functions between groups respect inverses and identity elements, as this will follow from the fact that they are multiplicative.

It seems unlikely that you already know the Isomorphism Theorems but you have not seen this little result, which is almost always established very early in the study of group homomorphisms, unless you know the Isomorphism Theorems from the general context already.

For the problem at hand, you are correct that if you can show this map $f\colon\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ is (i) a homomorphism; (ii) surjective); and (has kernel precisely $\langle (1,1)\rangle$; then you will be able to conclude that $(\mathbb{Z}\times\mathbb{Z})/\langle (1,1)\rangle$ is isomorphic to $\mathbb{Z}$. I don't know how you established the equality of the kernel with $\langle (1,1)\rangle$ (you need to prove both that everything in the kernel is a multiple of $(1,1)$ and that the subgroup is contained in the kernel). But assuming you've done it, and you've proven that $f$ is additive (the groups here are additive), then there is no need to establish separately that $f(-g) = -f(g)$ (note that these are additive groups, so $g^{-1}$ should be interpreted as "the additive inverse", not the reciprocal or multiplicative inverse).