7
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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
5
$\begingroup$

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 $\phi^m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.