# Why is $\mathbb{Z}$ not an inital object of Gr or AB?

Why is $$\mathbb{Z}$$ not an inital object of GR or AB?

Claim 1: for every group $$G$$ there exists a groups morphism from $$\mathbb{Z}$$ to $$G$$.

PF: Let $$f:\mathbb{Z} \rightarrow G$$ be given by: $$f(n) = n*1_G$$. Clearly $$f(1) = 1_G$$. Now $$f(n+m) = (n+m)*1_G = n*1_G + m*1_G$$. Hence $$f$$ is a groups hom.

Claim 2: $$f$$ is unique.

PF: This follows from that $$\mathbb{Z}$$ is generated by $$1$$ and hence every group hom starting from $$\mathbb{Z}$$ is completely determined by the image of $$1$$. And this image has to be $$1_G$$ by definition of group hom.

Now something is fishy here, because the trivial group is supposed to be the initial object of GR. And initial objects, when they exsist are unique.

• At some point you write $n \ast 1_G + m \ast 1_G$. What does this mean? If $G$ is a group, it carries only one operation. Did you mean to talk about (commutative) rings instead? Mar 21 '19 at 13:56
• $\mathbb{Z}$ is a group for the addition, not for the multiplication. Hence $1$ is not the neutral element and does not necessarily have to be mapped to $1_G$. Mar 21 '19 at 13:57

You've mixed up the additive and multiplicative identity of $$\mathbb{Z}$$. $$(\mathbb{Z},+)$$ is a group with identity $$0$$. A group homomorphism from $$\mathbb{Z}$$ must take the additive identity $$0$$ to $$1_G$$, but this does not determine the homomorphism. There are many morphisms from $$\mathbb{Z}$$ to a given group, in fact, mapping $$1 \in \mathbb{Z}$$ to any $$g \in G$$ defines a homomorphism.
The initial and final objects of $$\mathbf{Gr}$$ and $$\mathbf{Ab}$$ are the trivial group $$\{0\}$$.