Lagmental Vicfred

The Torsion Subgroup Is Functorial by Vicfred

Elements of finite order form a subgroup in every abelian group, and homomorphisms carry torsion to torsion. I want the notation, the mechanism, and the failure mode visible at the same time.

The data

Finite abelian groups become transparent after decomposing into \(p\)-primary components. For a cyclic group \(C_n\), element orders are controlled by \(\gcd(k,n)\) and direct products by least common multiples.

$$ T(A)=\{a\in A:\exists n\ge1,\ na=0\} $$

The typography mirrors the proof: first declare \(\mathsf D\), then state \(\mathsf C\). The symbol \(\Longrightarrow\) below is a logical dependency, not extra mathematical structure.

$$ f:A\to B\Longrightarrow f(T(A))\subseteq T(B) $$

Derivation

Here is a concrete symbolic test. Reading it from left to right reveals which transformation is reversible and which is only an implication.

$$ \begin{aligned}T(\mathbf Z^r\oplus C_{n_1}\oplus\cdots\oplus C_{n_s})&\cong\bigoplus_{i=1}^{s}C_{n_i},\\(\mathbf Z^r\oplus T)/T&\cong\mathbf Z^r.\end{aligned} $$

Invariant content

The formula is reusable precisely because it says which pieces are structural and which belong only to the worked example.

$$ \begin{aligned} \mathsf{D}\;&:\quad T(A)=\{a\in A:\exists n\ge1,\ na=0\},\\[5pt] \mathsf{C}\;&:\quad f:A\to B\Longrightarrow f(T(A))\subseteq T(B). \end{aligned} $$

Scope

An invariant such as order, exponent, or rank can rule out an isomorphism, but matching one invariant never proves two groups are isomorphic.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] f:A\to B\Longrightarrow f(T(A))\subseteq T(B) \end{gathered}} $$

The notation is dense, but it is doing honest work: every delimiter records scope and every index records dependence. Removing one should require a mathematical reason.

This article was posted on Thu 11 November 2021. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.