Question about finite abelian group Let G be an abelian group of order $mn$ where $\gcd(m,n)=1$.
I proved that $mG$ and $nG$ are subgroups and that $G=mG+nG$ and now i want to prove the three things:


*

*the sum is direct, i.e. $mG\cap nG=0$

*$nG$ has order $m$


For the first i take $x=mg_1=ng_2$. how do i show that $x=0$?
how can i show 2.? i'm really lost.
 A: The set $kG$, for an integer $k$, is the image of the endomorphism $x\mapsto kx$, so it is a subgroup.
By Bézout's theorem, $1=mr+ns$, for some integers $r$ and $s$; if $x\in G$, then $x=1x=(mr+ns)x=m(rx)+n(sx)\in mG+nG$.
Consider the surjective homomorphism $\mu_m\colon G\to mG$ defined by $x\mapsto mx$; then clearly $nG\subseteq\ker\mu_m$, so $|mG|$ divides $|G/nG|=|G|/|nG|$ and, consequently, $|mG|\cdot|nG|$ divides $|G|$.
Consider the homomorphism $\varphi\colon mG\oplus nG\to G$ defined by $\varphi(x,y)=x+y$. From $mG+nG=G$ it follows that $\varphi$ is surjective. Since $|mG\oplus nG|=|mG|\cdot|nG|\le |G|$, it follows that $|mG|\cdot|nG|=|G|$ and that $\varphi$ is also injective.
Now, if $x\in mG\cap nG$, we have $\varphi(x,-x)=x-x=0$, so $(x,-x)=(0,0)$ by injectivity of $\varphi$.
In order to show that $|mG|=n$ we just need to show that $|mG|$ divides $n$, because the same proof will show that $|nG| divides $m$.
Note that $nG=\{x\in G:mx=0\}$. Indeed, if $x=ny\in nG$, we have $mx=mny=0$. Conversely, if $mx=0$, write $x=1x=mrx+nsx=nxs\in nG$.
Let $p^r$ be a prime power dividing $|mG|$, with $r\ge1$. Then $mG$ has an element $x$ of order $p$. If $p\mid m$, we have $m=pt$ and $mx=ptx=0$, so $x\in nG$. Then $x\in mG\cap nG=\{0\}$, a contradiction. So any prime power dividing $|mG|$ divides $n$, and therefore $|mG|$ divides $n$.
