Annihilator - Product of cyclic groups

Question

Let $M$ be the abelian group, i.e., a $\mathbb{Z}$-module, $M=\mathbb{Z}_{24}\times\mathbb{Z}_{15}\times\mathbb{Z}_{50}$.

I want to find the annihilator $\text{Ann}(M)$ in $\mathbb{Z}$.

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$$\text{Ann}_{\mathbb{Z}}(M)=\{z\in \mathbb{Z}\mid \forall m\in M, zm=0\}$$

We have that $m\in M : m=(m_1, m_2, m_3)$, where $m_1\in \mathbb{Z}_{24}, m_2\in \mathbb{Z}_{15}, m_3\in \mathbb{Z}_{50}$.

Therefore, $zm=0\Rightarrow (zm_1, zm_2, zm_3)=0$.

$z$ must be the common multiples of $24, 15, 50$, right?

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I want to find also for the ideal $I=2\mathbb{Z}$ of $\mathbb{Z}$ the $\{m\in M\mid am=0, \forall a\in I\}$ as a product of cyclic groups.

Could you give me some hints how we could do that?

Answer 1

Hints:

For you second question, note that if $M=\oplus_k M_k$, then (valid in any commutative ring) $$(0:I)_M=\bigoplus_k(0:I)_{M_k}$$

Furthermore, $$(0:r)_{\mathbf Z/n\mathbf Z}=\begin{cases}d\mathbf Z/n\mathbf Z&\text{if}\enspace r\mid n,\enspace n=rd,\\0&\text{if}\enspace r\wedge n=1.\end{cases} $$