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Let $A = \text{Tor}_p(A)$ be a finitely generated abelian $p$-group. (Here $p$ is prime). Show that the minimal number of generators of $A$ is $\log_p|A/pA|$.

What I tried -

I think that from the fundamental theorem of finitely generated abelian groups A is isomorphic to a group of the form $\mathbb Z_p\oplus\mathbb Z_p \oplus \cdots$, so it can be represented as an $n$-dimensional vector space.

I'm not sure if it is right and how to continue from here.

Any help will be appreciated.

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    $\begingroup$ $\mathbb{Z}/(4)$ is also a $2$-group. $\endgroup$
    – Loki Clock
    Commented Dec 28, 2015 at 11:01

1 Answer 1

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A finitely generated abelian $p$-group $A$ looks like $$\mathbb Z/p^{k_1}\mathbb Z\oplus\cdots\oplus\mathbb Z/p^{k_s}\mathbb Z$$ with $k_1\ge\cdots\ge k_s\ge 1$. Then $$A/pA\simeq(\mathbb Z/p\mathbb Z)^s,$$ so $|A/pA|=p^s$, and therefore $\log_p|A/pA|=s$.

Now I leave you the pleasure to find out why $A$ can't have less than $s$ generators.

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  • $\begingroup$ still not sure about the generators.. Is it because A represent an s dimensional vector space, so it is not possible that it has less than s generators?? thanks. $\endgroup$
    – user101010
    Commented Dec 29, 2015 at 20:57
  • $\begingroup$ @gpgpgp $A/pA$ is an $s$-dimensional vector space over $\mathbb Z/p\mathbb Z$. What's going on if $A$ has less than $s$ generators? $\endgroup$
    – user26857
    Commented Dec 29, 2015 at 21:07

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