Let $G$ be a finite abelian group and let $\phi: G \rightarrow G$ be a group homomorphism. I am trying to show that there is a positive integer $n$ such that $G \cong \ker(\phi^{n}) \times \phi^{n}(G)$.

I know that since $G$ is abelian we have that $\ker(\phi^{k})$ and $\phi^{k}(G)$ are normal subgroups of $G$ for any $k$

I also suspect we have the following towers which stabilize at $m$ and $m'$

$$\phi(G) \unrhd \phi^{2}(G) \unrhd \cdots \unrhd \phi^{m}(G)=\phi^{m+1}(G)=\cdots$$

$$\ker(\phi) \unlhd \ker^{2}(\phi) \unlhd \cdots \unlhd \ker^{m'}(\phi)=\ker^{m'+1}(\phi)=\cdots$$

I know that if given a group of the form $HK$ where $H$ and $K$ are normal subgroups of $HK$ and $H\cap K=1$ then $H \times K \cong HK$. I want to apply this to this situation but I am unable to show how write $G$ as a product $HK$

Resolution One of the comments directed me to the following article: http://en.wikipedia.org/wiki/Fitting_lemma The last part of which answers the question. Below is what is says:

Choose $n=\max(m,m')$, then we have for $x \in \ker^{n}(\phi) \cap \phi^{n}(x)$, this means that $x=\phi^{n}(y)$ for some $y \in G$. This gives:

$0=\phi^{n}(x)=\phi^{2n}(y)$ which means that $y \in \ker^{2n}\phi=\ker^{n}{\phi}$. and then we have that $0=x=\phi^{n}(y)$.

The answer below kindly points out that every element $x$ is contained in one of the cosets of $G/\ker^{n}(\phi)$, this means that $x=k+g$ for some $k \in \ker^{n}(\phi)$ and $g \in G$. We also have that $g=\phi^{n}(h)$ for some $h \in G$. Writing $x=k+\phi^{n}(h)$ essentially shows that $G=\ker(\phi^{n}) + \phi^{n}(G)$. Using the fact above, the claim follows.

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    $\begingroup$ See en.wikipedia.org/wiki/Fitting_lemma ("Consequently, $M$ is the direct sum of $\operatorname{im}\left(f^k\right)$ and $\operatorname{ker}\left(f^k\right)$"). $\endgroup$ – darij grinberg Dec 23 '14 at 20:17
  • $\begingroup$ Thanks, I think the last part of the article answers my question. I will make an edit including what the article says soon. Thanks a lot. $\endgroup$ – user135520 Dec 23 '14 at 20:43
  • $\begingroup$ Just saw your comment after I posted my answer; the phone app doesn't let you see activity while you're typing an answer. Hope the answer isn't too offensive. $\endgroup$ – Matt Samuel Dec 23 '14 at 20:48
  • $\begingroup$ No, not at all. Thank you $\endgroup$ – user135520 Dec 23 '14 at 20:49

Suppose the image of $\phi^m$ is the same as the image of $\phi^{m+1}$. Then the kernels are also the same. Every element of the group is contained in some coset of the kernel of $\phi^m$, so every element $g$ is equal to $\phi^m(h)+k$, where $h\in G$ and $k$ is in the kernel. $\phi^m$ is an automorphism when restricted to its image because the image has stabilized. Thus the kernel has trivial intersection with the image. These facts, as you have noted, show that $G$ is the direct sum of the kernel and the image of $\phi^m$.


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