Show how if $\gcd(m,n)=1$, $\mathbb{Z}_{mn}$ is the internal direct product of two subgroups.

I know that internal direct products must meet three criteria.

  1. $G=HK=\{hk:h\in H, k\in K\}$
  2. $hk=kh$ for all $h\in H$ and $k\in K$
  3. $H\cap K=\{e\}$ (the identity).

I'm struggling with relating this to the $\gcd(m,n)=1$.

  • 2
    $\begingroup$ I think the first step should be identifying the two subgroups. $\endgroup$ – asdq Feb 1 '18 at 23:12

You probably can start by trying $\langle\bar m\rangle \times \langle \bar n\rangle$. Then since $\mathbb{Z}_{mn}$ is abelian, your criterion (2) is automatically fulfilled.

Criteria (1) and (3) come from $lcm(m,n)=mn$ and $gcd(m,n)=1$ respectively. So your assumption on $gcd$ is essential.

Consider $\mathbb{Z}_4$ then you cannot write it as $\mathbb{Z}_2\times\mathbb{Z}_2$ ($\mathbb{Z}_4$ is cyclic, i.e. generated by an element $\mathbb{Z}_4=\langle 1\rangle$, while $\mathbb{Z}_2\times\mathbb{Z}_2$ is not; also $\mathbb{Z}_4$ has elements of order 4 (1 and 3), but $\mathbb{Z}_2\times\mathbb{Z}_2$ does not [all non-identity element has order 2]) (Compare with $\mathbb{Z}_6\cong\mathbb{Z}_2\times\mathbb{Z}_3$ by Chinese remainder theorem. )

  • $\begingroup$ Didn't you mean $\mathbb{Z}/(m\mathbb{Z}) \times \mathbb{Z}/(n\mathbb{Z})$? $\endgroup$ – Stefan4024 Feb 1 '18 at 23:57
  • 1
    $\begingroup$ I do not know how to articulate my notations well, but take an example of $\mathbb{Z}_6$ (here $6=2\times 3$), then I meant the two subgroups are $\langle 2\rangle$ and $\langle 3\rangle$; or, explicitly, $\{\bar 0,\bar 2,\bar 4\}$ and $\{\bar 0,\bar 3\}$ respectively. $\endgroup$ – ElfHog Feb 1 '18 at 23:59
  • $\begingroup$ One way to write it would be $\frac{m\mathbb{Z}}{mn\mathbb{Z}}$ and $\frac{n\mathbb{Z}}{mn\mathbb{Z}}$ which are isomorphic to $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}/m\mathbb{Z}$, resp. $\endgroup$ – André 3000 Feb 2 '18 at 0:36
  • $\begingroup$ @ElfHog In general $m\mathbb{Z} = \{mn | n \in \mathbb{Z}\}$, because these sets denote cosets in $\mathbb{Z}$, so they aren't related to the subgroup $\mathbb{Z}_{mn}$ directly. You have to use factor groups. $\endgroup$ – Stefan4024 Feb 2 '18 at 0:50
  • $\begingroup$ Thanks for commenting. I guess the notation now is better. Please tell me if my notations are still sloppy. $\endgroup$ – ElfHog Feb 2 '18 at 0:59

Since $m\mathbb{Z},mn\mathbb{Z}$ are normal subgroups of $\mathbb{Z}$ and $mn\mathbb{Z}\subset m\mathbb{Z}$, we have that $m\mathbb{Z}/mn\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}/mn\mathbb{Z}$. Similarly, $n\mathbb{Z}/mn\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}/mn\mathbb{Z}$.

We show that $m\mathbb{Z}/mn\mathbb{Z}\cap n\mathbb{Z}/mn\mathbb{Z}=0$.

Let $ma+mn\mathbb{Z}=nb+mn\mathbb{Z}\in m\mathbb{Z}/mn\mathbb{Z}\cap n\mathbb{Z}/mn\mathbb{Z}$, where $a,b\in \mathbb{Z}$. Then $ma-nb\in mn\mathbb{Z}$. So $ma-nb=mnc$ for some $c\in \mathbb{Z}$. Hence $m|nb$. Since $(m,n)=1$, $m|b$. So $b=md$ for some $d\in \mathbb{Z}$. Hence $nb+mn\mathbb{Z}=mnd+mn\mathbb{Z}=0$.

We show that $\mathbb{Z}/mn\mathbb{Z}=(m\mathbb{Z}/mn\mathbb{Z})(n\mathbb{Z}/mn\mathbb{Z})$.

Since $(\mathbb{Z}/mn\mathbb{Z})/(m\mathbb{Z}/mn\mathbb{Z})\cong \mathbb{Z}/m\mathbb{Z}$, $|m\mathbb{Z}/mn\mathbb{Z}|=mn/m=n$. Similarly, $|n\mathbb{Z}/mn\mathbb{Z}|=m$. Then $$|(m\mathbb{Z}/mn\mathbb{Z})(n\mathbb{Z}/mn\mathbb{Z})|=|m\mathbb{Z}/mn\mathbb{Z}|\cdot|n\mathbb{Z}/mn\mathbb{Z}|/|m\mathbb{Z}/mn\mathbb{Z}\cap n\mathbb{Z}/mn\mathbb{Z}|=mn.$$ Since $|\mathbb{Z}/mn\mathbb{Z}|=mn$, we have $\mathbb{Z}/mn\mathbb{Z}=(m\mathbb{Z}/mn\mathbb{Z})(n\mathbb{Z}/mn\mathbb{Z})$.


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