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
    Mar 21 '19 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
    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 $\{0\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.