Problem reagarding the annihilator of a $\mathbb Z$-module

Question

Let $M=\mathbb Z/24\mathbb Z\times \mathbb Z/15\mathbb Z\times \mathbb Z/50\mathbb Z$.
a) Find the annihilator of $M$ in $\mathbb Z$.
b) Let $I=2\mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.

Annihilator of $M$ is $\{r\in \mathbb Z|rm=0\forall m\in M\}$.

My attempt:

a) Annihilator of $M=d\mathbb Z$ where $d=\gcd(24,50,15)$.

b) Actually I do not get this question. I would be thankful if someone help me. Please give me hint.

Any help will be appreciated.

Answer 1

a) By your work, $\gcd(24,50,15)=1$ and the annihilator is $1\mathbb Z=\mathbb Z$. Does that sound right? Perhaps instead of fishing for an answer that sounds plausible, you should think of at least one element you know can annihilate $M$, and then reconsider what you've written.

b)

Let $I=2\mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.

By definition, the annihilator of $I$ in $M$ is $\{m\in M\mid Im=0\}$. This is what you should be looking for. Locate these elements and see if you can decompose them as a product of cyclic groups. Knowing it is a subgroup of $M$ should help you a lot.

Good luck.