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I am trying to solve the following problem:

Determine which abelian groups $A$ can appear as central terms in a short exact sequence $\mathbb{Z} \to A \to \mathbb{Z} \oplus \mathbb{Z}_5$

What I've done so far is the following:

Calling the first map $i$ and the second $\pi$ I know that im($i$)=ker($\pi$) so if I call $i(\mathbb{Z})=H \subset A$ I know that

$$ \mathbb{Z} \oplus \mathbb{Z}_5 \cong A/H $$

This condition is clearly satisfied considering $A = \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_5$ but I don't know how to determine if this is the only group which makes the sequence exact or if there are more.

I thought of trying to get conditions to know if the sequence splits or not but I haven't been able to get far.

How should I proceed?

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  • $\begingroup$ I suppose $\mathbb{Z}_5$ denotes the cyclic group of order $5$ and not the ring of $5$-adic integers? $\endgroup$ Jan 16, 2016 at 12:01
  • $\begingroup$ @DanielFischer Yes, it is the cyclic group of order 5. $\endgroup$
    – S -
    Jan 16, 2016 at 12:02
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    $\begingroup$ Okay. Then $\mathbb{Z}_5 = \mathbb{Z}/(5\mathbb{Z})$ should give you an idea for another possibility for $A$. $\endgroup$ Jan 16, 2016 at 12:03
  • $\begingroup$ Do you know the structure theorem for finitely generated abelian groups? $\endgroup$ Jan 16, 2016 at 12:05
  • $\begingroup$ @DanielFischer Yes, my teacher explained it to the class a couple of lessons ago $\endgroup$
    – S -
    Jan 16, 2016 at 12:06

1 Answer 1

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Thanks to @DanielFischer comment I was able to get the answer by myself.

Considering $i(1)=(5,0)$ then the $A= \mathbb{Z} \oplus \mathbb{Z}$ is another valid option. These are the two only valid options. Let's see the intuition behind this.

$A = LT(A) \oplus T(A)$ where $T(A)$ is the torsion part of $A$. We can racionalize the sequence and taking into account that racionalizing preserves exact sequences we get:

$$ \mathbb{Q} \to A_\mathbb{Q} \to \mathbb{Q}$$

The exactness of this sequence gives us that the rank of A must be equal to the rank of $\mathbb{Z} \oplus \mathbb{Z}$.

If the torsion part of $A$ is 0 we have to generate it when taking the quotient with respect to $i(\mathbb{Z})$ this allows us to prove that if $i(1) \neq 5$ we don't get $\mathbb{Z}_5$.

On the other hand, if $T(A) \neq 0$ then when taking the quotient with respect to $\mathbb{Z}$ we get the first group proposed in the answer as the only chance.

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  • $\begingroup$ Great. I love it when a little nudging works. $\endgroup$ Jan 16, 2016 at 12:24

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