I am trying to understand the following examples in more details. Why are the following hold?? Can we prove it in details?
- Let the class $$Y = \{Z_n:n=2,3,\cdots\},$$ then for each abelion group, the trace $\mathrm{tr}(Y)$ over $M$ is the torsion subgroup $T(M)$ of $M$. Also, $T(M)$ is the unique largest torsion subgroup of $M$, and $T(T(M))=T(M)$.
- If $M$ is an abelion group, then $\mathrm{rej}(Q)$ over $M$ is the intersection of all $K$ subgroup of $M/K$ torsion free. So $\mathrm{rej}(Q)$ over $M$ is just the torsion subgroup $T(M)$ of $M$, the unique smallest subgroup with $M/T(M)$ torsion free. And $T(M/T(M))=0$.
Let $M$ be an abelian group. If $x\in M$ has finite order $n$, then there exists $f\colon\mathbb{Z}_n\to M$ such that $x$ belongs to the image of $f$.
Hence every torsion element of $M$ belongs to the trace of $\mathcal{Y}$. In particular, $T(M)\subseteq\operatorname{tr}(\mathcal{Y})$. The converse is easy.
Consider a homomorphism $f\colon M\to\mathbb{Q}$. If $x\in T(M)$, then $f(x)$ is torsion as well, so $f(x)=0$. Hence $T(M)\subseteq\ker f$ and consequently, $T(M)\subseteq\operatorname{rej}(\mathbb{Q})$.
Since $M/T(M)$ is torsion free (prove it), all you need to show now is that, if $N$ is a torsion free group and $x\in N$, $x\ne0$, then there exists $f\colon N\to\mathbb{Q}$ such that $f(x)\ne0$. This relies on the fact that $\mathbb{Q}$ is divisible, hence injective as a $\mathbb{Z}$-module.