When is $n\mathbb{Z}_{mn}$ a summand of $\mathbb{Z}_{mn}$?

Question

I have done this for different examples like is $2\mathbb{Z}$ a summand of $\mathbb{Z}$?(the answer is no in this case) and is $2\mathbb{Z}_{4}$ a summand of $\mathbb{Z}_{4}$?(the answer is yes in this case). But what about in the more general case of $\mathbb{Z}_{mn}$? Any hints/tips would be appreciated?

Answer 1

It wil be true if and only if $n,m$ are coprime integers, which confirms the fact that $2\mathbb{Z}/4\mathbb{Z}$ is NOT a direct summand of $\mathbb{Z}/4\mathbb{Z}$ (cf. Arturo's comment).

Set $R=\mathbb{Z}/nm\mathbb{Z}$. If $n\cdot R\oplus L=R$, then $L\simeq R/n\cdot R\simeq \mathbb{Z}/n\mathbb{Z}$ (as group. Left as an exercise).

Hence $L$ is in particular a cyclic subgroup of $R$ of order $n$, that is $L=m\cdot R$.This is also an $R$-module.

Now, $n\cdot R\cap m\cdot R=lcm(n,m)\cdot R$, which will be $0$ if and only if $lcm(n;m)=nm$, that is if and only if $n,m$ are coprime (details are left to you).

Conversely, if $n,m$ are coprime, you have $nR\oplus mR=R$: we have $nR\cap mR=0$, and if $u,v\in\mathbb{Z}$ satisfy $un+vm=1$, then for all $\bar{x}\in R$, we have $\overline{x}=n\cdot \overline{ux}+m\cdot \overline{vx}.$