1
$\begingroup$

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 $\gcd(m,n)=1$.

$\endgroup$
  • 2
    $\begingroup$ I think the first step should be identifying the two subgroups. $\endgroup$ – asdq Feb 1 '18 at 23:12
0
$\begingroup$

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. )

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.