# Endomorphisms of a finitely generated abelian group

• Sep 6th 2011, 09:31 AM
NonCommAlg
Endomorphisms of a finitely generated abelian group
I think this is a nice problem:

Problem. Explicitly describe the (ring of) endomorphisms of a finitely generated abelian group $G.$

The answer should be in a matrix form.
• Sep 7th 2011, 04:56 AM
Swlabr
Re: Endomorphisms of a finitely generated abelian group
Well, almost by definition the set of $m\times m$ matrices with entries from $\mathbb{Z}_{n^a}$ form the endomorphism ring of $\mathbb{Z}_{n^a}^m$ (with $n\in\mathbb{N}\cup\{\infty\}$, $\mathbb{Z}_{\infty}:=\mathbb{Z}$). So the question is "how do we pin these together?" Basically, you pin them together by assuming you have your fgab in a "collected" form, with different copies of isomorphic groups beside each other,

$G\cong \mathbb{Z}_{n_1^{a_1}}^{m_1}\times\mathbb{Z}_{n_2^ {a_2}}^{m_2}\times\ldots\times\mathbb{Z}_{n_p^{a_p }}^{m_p}$ where $\mathbb{Z}_{n_i^{a_i}}$ appears nowhere else in the decomposition of your group.

Then, your ring of endomorphisms is the ring of $m_1+m_2+\ldots+m_p$ matrices, where the top $m_1$ elements on the diagonal, and the $m_1\times m_1$ square around them, have entries from $\mathbb{Z}_{n_1^{a_1}}$, the $(m_1+1)^{th}$ to the $m_1+m_2$ elements on the diagonal, and the $m_2\times m_2$ square around them have entries from $\mathbb{Z}_{n_2^{a_2}}$, etc. Everywhere else is zero.

Clearly all such matrices form a ring, and clearly they are endomorphisms of the abelian group. It is then sufficient to prove that every endomorphism appears in this way, but that is routine...and would take too long to type (I'm lazy).
• Sep 8th 2011, 09:40 AM
NonCommAlg
Re: Endomorphisms of a finitely generated abelian group
yes, we first use the fundamental theorem for finitely generated abelian groups and then, for abelian groups $G_i,$ we apply the ring isomorphism $\text{End} \left (\bigoplus_{i=1}^n G_i \right ) \cong R,$ where $R$ is the matrix ring whose $(i,j)$-entry is $\text{Hom}(G_j,G_i).$ so we just need to find $\text{Hom}(G_j,G_i),$ where $G_i$ and $G_j$ are cyclic groups (finite or infinite). that's the idea but the full solution is quite long.