7
$\begingroup$

This seemingly simple question has puzzled me for a while: Determine which abelian groups A fit into a short exact sequence $$0 \to \mathbb{Z}_{p^m} \to A \to \mathbb{Z}_{p^n} \to 0$$

(This is question 2.1.14 in Hatcher's Algebraic Topology)

First and foremost, I'd like to see an elementary straightforward solution. If there is a more sophisticated solution, I'd be happy to see it as well (one approach we had in mind is to compute the value of Ext, which didn't work quite well).

  • 1
    Computing $\operatorname{Ext}^1(\mathbb Z_{p^n},\mathbb Z_{p^m})$ is somewhat straight forward, just take a free resolution $\mathbb 0\to Z \to \mathbb Z \to \mathbb Z_{p^n}$ and apply $\hom$ (I may be swapping $m$ and $n$), but it doesn't quite give you the groups $A$. For every element of $\operatorname{Ext}$ you can construct an extension in a somewhat algorithmic way, but there is definitely an extra step that needs to be done, which is not nearly as well known as the basic construction of $\operatorname{Ext}$ is. If I have time later, I will say something about it.2011-05-31
  • 1
    This should be pretty straightforward using the structure theorem, which will at least tell you what the answer is, and then you can try to figure it out more directly.2011-05-31
  • 0
    Oh, oops, I missed the fact that $A$ is abelian. That makes things somewhat simpler. In that case, using $\operatorname{Ext}$ is definitely NOT the way to go. As Qiaochu says, the structure theorem for finitely generated abelian groups is a much better starting point.2011-05-31
  • 0
    @Qiaochu: Obviously, the size of A must be p^(n+m), so by the structure theorem we know that A is a sum of cyclic groups of size p^(a_i) such that the sum of the a_i's is m+n. This still leaves quite some room for different decompositions, and it's not clear which decomposition fits in the series.2011-05-31
  • 0
    Isn't it? Given any particular sequence $a_i$ you can tell whether it has a subgroup isomorphic to $\mathbb{Z}/p^m\mathbb{Z}$, and having found such a subgroup you can then tell whether the quotient is isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$. This requires a little work with one's hands but I don't see anything conceptually difficult about it.2011-05-31
  • 0
    @Aaron: why isn't Ext the way to when A is abelian? In fact, it was when you thought that A need not be abelian that Ext would not work :)2011-06-01
  • 0
    @Mariano Suárez-Alvarez Oops, you're right. I was jumbling a few things in my head when I wrote that.2011-06-01
  • 2
    There is a solution in [www.math.ku.dk/~moller/f03/algtop/opg/S2.1.pdf](http://www.math.ku.dk/~moller/f03/algtop/opg/S2.1.pdf), although I don't fully understand the argument. Those with better algebra skills then me probably will though2011-06-01

2 Answers 2