MHB How to Find the Product of Cyclic Groups in an Abelian Group?

[mathmari]

Hey!

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 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.

We have the following:

$$\text{Ann}_M(I)=\{m\in M\mid am=0, \forall a\in I\} \\ =\{m\in \mathbb{Z}_{24}\mid am=0, \forall a\in I\}\times\{m\in \mathbb{Z}_{15}\mid am=0, \forall a\in I\}\times\{m\in \mathbb{Z}_{50}\mid am=0, \forall a\in I\}\\ =\text{Ann}_{\mathbb{Z}_{24}}(I)\times\text{Ann}_{\mathbb{Z}_{15}}(I)\times\text{Ann}_{\mathbb{Z}_{50}}(I)$$

or not? (Wondering)

 

[Euge]

It looks good so far. :D

 

[mathmari]

It looks good so far. :D

How could we continue? Do we have to show that $\text{Ann}_{\mathbb{Z}_{24}}(I), \text{Ann}_{\mathbb{Z}_{15}}(I), \text{Ann}_{\mathbb{Z}_{50}}(I)$ are cyclic groups? (Wondering)

 

[Euge]

They are cyclic, but based on your prompt I thought you have to compute them. I'll start with $\operatorname{Ann}_{\Bbb Z_{24}}(I)$. Since $2$ is a generator of $I$, an element $m\in \Bbb Z_{24}$ satisfies $am = 0$ if and only if $2m = 0$. The solutions of $2m = 0$ in $\Bbb Z_{24}$ are $[0]_{24}$ and $[12]_{24}$. Thus $\operatorname{Ann}_{\Bbb Z_{24}}(I) = 12\Bbb Z_{24}$.

 

[mathmari]

They are cyclic, but based on your prompt I thought you have to compute them. I'll start with $\operatorname{Ann}_{\Bbb Z_{24}}(I)$. Since $2$ is a generator of $I$, an element $m\in \Bbb Z_{24}$ satisfies $am = 0$ if and only if $2m = 0$. The solutions of $2m = 0$ in $\Bbb Z_{24}$ are $[0]_{24}$ and $[12]_{24}$. Thus $\operatorname{Ann}_{\Bbb Z_{24}}(I) = 12\Bbb Z_{24}$.

Ah ok...I see... (Thinking)

So, we have also the following:

Since $2$ is a generator of $I$, an element $m\in \Bbb Z_{15}$ satisfies $am = 0$ if and only if $2m = 0$. The solution of $2m = 0$ in $\Bbb Z_{15}$ is $[0]_{15}$. Thus $\operatorname{Ann}_{\Bbb Z_{15}}(I) = 15\Bbb Z_{15}$.

Since $2$ is a generator of $I$, an element $m\in \Bbb Z_{50}$ satisfies $am = 0$ if and only if $2m = 0$. The solutions of $2m = 0$ in $\Bbb Z_{50}$ are $[0]_{50}$ and $[25]_{50}$. Thus $\operatorname{Ann}_{\Bbb Z_{50}}(I) = 25\Bbb Z_{50}$.

Is this correct? (Wondering)

 

[Euge]

Yes, it's correct.