An epimorphism of $G=H \oplus \Bbb Z^N$ onto $G$, where $H$ is a finite abelian group Let $H$ be an abelian group with $|H|<\infty$, and let $N$ be a positive integer, and consider the group $G=H \oplus \Bbb Z^N$. 


*

*If $f$ is a surjective endomorphism of $G$ then is $f$ an automorphism?

*Similarly, if $f$ is an injective endomorphism of $G$ then is $f$ an automorphism?
How do I have to proceed? I even can't see that these statements are true. 
 A: So for (2), consider $N=1$ and $(h, z)\mapsto(h, 2z)$. This is injective, but not surjective - the answer therefore is negative. The first part, however, is true. To see this, first note that $K:=\ker(f)$ only contains elements of the form $(h,0)$. Indeed, if $(h,v)\in K$ for some $v\ne 0$, then by multiplying with $m:=|H|$ we may fist assume that $h=0$ (this just makes every entry of $v$ larger). Then, $\{0\}\oplus \Bbb Zv \subseteq K$, so $K$ contains a subgroup that is isomorphic to $\Bbb Z$. However, this implies that the rank of $K$ is at least one. Since
$$
0\longrightarrow K\hookrightarrow H\oplus \Bbb Z^N\xrightarrow{~~~f~~~} H\oplus \Bbb Z^N\longrightarrow 0
$$
is exact and the rank is additive, this yields a contradiction. We also note that $f(h,0)\subseteq H\oplus\{0\}$. Indeed, if $(x,y)=f(h,0)$ and $m:=|H|$, then 
$$
(0,0)= f(mh,0)=mf(h,0)=(mx,my)
$$
and so $0=my$, implying $y=0$. This defines an endomorphism $\varphi:H\to H$ with $f(h,0)=(\varphi(h),0)$. Now assume for contradiction that $\ker(\varphi)\oplus\{0\}=K\ne\{0\}$, i.e. $\ker(\varphi)\ne\{0\}$. Then we know that $\varphi$ is not surjective and there is a $z\in H$ which is not in the image. However, there is a tuple $(h, v)$ such that $(z,0)=f(h,v)=(\varphi(h),0)+f(0,v)$. It follows that $f(0,v)=(h',0)$ where $h'=z-\varphi(h)\ne 0$. In particular, $v\ne 0$ and $f(0, mv)=0$, so $m\Bbb Z v\subseteq K$, which contradicts that $K$ is finite.
