Question on why the isomorphism $A \cong \mathbf{Z}^n \oplus \text{Tor} A$ is not natural -- A clarification of Riehl's choice of group?

Question

I'm reading Category Theory in Context, and I have a clarification question.

Her Proposition 1.4.4 says that the isomorphism of f.g. Abelian groups $A \cong \mathbf{Z}^n \oplus \text{Tor} A$ is not natural. The argument goes as follows.

Lemma: All natural endomorphisms of the identity functor in $\mathsf{f.g.Group}$ have components $n(-)$ for fixed $n$.

Suppose that the above isomorphism is natural. Then the isomorphism $\alpha$:

$$A \to A \text{ } /\text{ Tor}A \cong \mathbf{Z}^n \hookrightarrow \mathbf{Z}^n \oplus \text{Tor} A \cong A$$

is natural as well. Since $\mathbf{Z}$ has zero torsion part, it follows that following the above natural isomorphism gives the component $\alpha_\mathbf{Z} = n(-)$ for nonzero $n$.

Riehl then takes $\mathbf{Z}/(2n)$, which is all torsion, to illustrate that $\alpha_{\mathbf{Z}/(2n)} = 0$, contradicting the lemma.

Here is my confusion: We could pick $\mathbf{Z}/(q)$ for any $q \neq 0$ for the last step of the argument, correct? The fact that the chosen group is dependent on $n$ makes me suspicious that $n$ is relevant here, but any finite cyclic group should work, correct? Any such group is entirely torsion, hence would have $0$ component. Did Riehl simply chose $\mathbf{Z}/(2n)$ arbitrarily, or was there a reason?

Answer 1

The argument isn't quite how you've presented it. Let's write $TA$ for the torsion part of $A$, as Riehl does. (I'm working from p. 26 of https://math.jhu.edu/~eriehl/context.pdf.)

First suppose that the isomorphism is natural. Then the endomorphism (not isomorphism) $$ A \to A/TA \to TA \oplus A/TA \cong A $$ defines a natural endomorphism $\alpha$ of the identity functor on the category of finitely generated abelian groups.

Lemma: every natural endomorphism of the identity functor is just multiplication by $n$ for some integer $n$.

Since the above map is nonzero when $A = \mathbb{Z}$, then the integer $n$ must be nonzero (looks to me like it must be $\pm 1$). For any torsion group, the above map is multiplication by the same integer $n$ but is also certainly zero, so we will get a contradiction if we find a torsion group in which $n \neq 0$. That's why she chose $\mathbb{Z}/(2n)$.