For $A=\mathbb{Z}_m$, show $A_n=\mathbb{Z}_d$ where $d=gcd(m,n)$.

Question

I am struggling some with proving the following. I am struggling the most with showing $A_n\subseteq\mathbb{Z}_d$.

Prove that $A_n=\{a|na=0\}$ where $n$ is fixed in $\mathbb{Z}$ is a submodule of $A$. For $A=\mathbb{Z}_m$, show that $A_n=\mathbb{Z}_d$, where $d=(m,m)$.

Below is what I have tried. Also if you could check what I have so far, it would be much appreciated. Thank you!

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Assume hypothesis. Clearly $A_n\subseteq A$ by definition. We want to show that $A_n$ is a submodule of $A$ by proving (1) $A_n\neq\emptyset$ and (2) $x+zy\in A_n$ for all $z\in\mathbb{Z}$ and for all $x,y\in A_n$.

1. Clearly $0\in A$ because 0 is the additive identity of $\mathbb{Z}$. Notice $n\cdot 0=0$. Therefore $0\in A_n$, which means $A_n\neq\emptyset$. This satisfies the 1st property.
2. Let $z\in \mathbb{Z}$ and $x,y\in A_n$. Then $nx=0$ and $ny=0$. We want to show $x+zy\in A_n$. Consider $$ n(x+zy)=nx+z(ny) = 0 + z\cdot 0 = 0. $$ Therefore $x+zy\in A_n$. This satisfies the 2nd property.

Because the two properties are satisfied, $A_n$ is a submodule of $A$.

Now we want to show that for $A=\mathbb{Z}_m$, we have $A_n=\mathbb{Z}_d$ where $d=\gcd(m,n)$.

$(\subseteq$) Let $x\in A_n$. Then $nx=0$. Then $x\in\mathbb{Z}_n$. Because $A_n$ is a submodule of $A$, we know $x\in A_n\subseteq A=\mathbb{Z}_m$. Therefore $x\in \mathbb{Z}_n\cap \mathbb{Z}_m$. Let $d=\gcd(m,n)$. Notice $\mathbb{Z}_d\subseteq\mathbb{Z}_m\cap\mathbb{Z}_n$.

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Question: Here is where I am getting a little stuck and am not sure how to exactly proceed. I feel like I am missing something obvious to conclude $x\in\mathbb{Z}_d$.

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($\supseteq$) Let $x\in \mathbb{Z}_d$. Notice $x\in[x]=\{x+zd|z\in\mathbb{Z}\}\in\mathbb{Z}_d$. By a theorem, we know $d=ms+nt$ such that $s,t\in \mathbb{Z}$. Then \begin{eqnarray} x+zd &=& x+z(ms+nt)\\ &=& x+ztn+zsm \end{eqnarray} This means $$x\in\{x+zd|z\in\mathbb{Z}\}=\{(x+ztn)+(zs)m|z,t,s\in\mathbb{Z}\}\in\mathbb{Z}_m=A$$ and $$x\in\{x+zd|z\in\mathbb{Z}\}=\{(x+zsm)+(zt)n|z,t,s\in\mathbb{Z}\}\in\mathbb{Z}_n.$$ Because $x\in\mathbb{Z}_n$, we know $nx=0$. Then because $x\in A$ and $nx=0$, we have $x\in A_n$ by definition.

Therefore $A_n\supseteq \mathbb{Z}_d$.

Hence for $A=\mathbb{Z}_m$, we have $A_n=\mathbb{Z}_d$ where $d=\gcd(m,n)$.

Answer 1

$\,\ na\!=\!0$ in $\,\Bbb Z_m\!\!\iff\! m\mid na\!\iff\! m\mid na,ma\!$ $\iff\! m\mid (na,ma)\! =\! \underbrace{(n,m)}_{d}a\!\iff\! m/d \mid a$

$\,\Bbb Z_m\,$ contains $d\,$ multiples of $\,c = m/d,\,$ viz. $\, 0,\, c,\, 2c,\,\ldots, (d\!-\!1)c,\,$ and it is easy to check that they form a subgroup isomorphic to $\,\Bbb Z_d.\,$