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.

  • 1
    $\begingroup$ 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? $\endgroup$
    – Bib-lost
    Commented Mar 21, 2019 at 13:56
  • 2
    $\begingroup$ $\mathbb{Z}$ 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 $1_G$. $\endgroup$
    – sTertooy
    Commented Mar 21, 2019 at 13:57

1 Answer 1


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 $\{0\}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .