# 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 $$G=H \oplus \Bbb Z^N$$.

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

2. 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.

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.