0
$\begingroup$

By rank $rk\ G$ of a group I mean the minimum set of its generators (even if the group is abelian, I consider it as if it were just a general group $-$ I do not mean here torsion-free rank!).

For a finite abelian group $A=Z_{p_1}^{m_1}\times\cdots\times Z_{p_n}^{m_n}$, where $Z_p=Z/pZ$ are cyclic groups and $p_i$ are prime numbers, is it true that $rk\ A=\sum m_i$? (Not torsion-free rank!)

In particular, is it true that for $A=Z_2^n=(Z/2Z)\times\cdots\times(Z/2Z)$ ($n$ copies), $rk\ A=n$?

If yes, could you please provide (or point to) a proof?

$\endgroup$
  • 2
    $\begingroup$ The minimal number of generators is not generally the sum of the $m_i$'s. For example, if each $m_i$ is 1 and the $p_i$ are distinct primes then $A$ is a cyclic group of order $p_1\cdots p_n$, so your rank for the group would be 1. If $A = ({\mathbf Z}/(2))^n$, or more generally if $A = ({\mathbf Z}/(p))^n$ for a prime $p$, then the minimal number of generators of $A$ is $n$, since this can be viewed as a linear algebra problem: additive generators of such $A$ are the same as a spanning set for $A$ as a vector space over ${\mathbf Z}/(p)$, so a minimal spanning set has size = dimension = $n$. $\endgroup$ – KCd Dec 27 '14 at 0:21
  • 1
    $\begingroup$ It might be of interest that for any $p$-group (even in the nonabelian case), the term "minimal generating set" conincides with "generating set of minimum cardinality", and the cardinality of any minimal generating set is equal to the dimension of $G/\Phi(G)$ over $\Bbb F_p$. It is a theorem that in this case $\Phi(G)=G'G^p$, so that $G/\Phi(G)$ is elementary abelian. $\endgroup$ – Pedro Tamaroff Dec 27 '14 at 0:26
  • 1
    $\begingroup$ Suppose $A = ({\mathbf Z}/(p))^a \times ({\mathbf Z}/(q))^b$ where $p$ and $q$ are distinct primes and $a \leq b$. Then $A = ({\mathbf Z}/(p))^a \times ({\mathbf Z}/(q))^a \times ({\mathbf Z}/(q))^{b-a} = ({\mathbf Z}/(pq))^a \times ({\mathbf Z}/(q))^{b-a}$, so $A$ has at least $a + b-a = b$ generators. If $m$ is the minimum number of generators of $A$ then $m \leq b$. Also $A/qA \cong (\mathbf Z/(q))^b$ has $m$ generators, so by linear algebra $m \geq b$. Thus your rank is $b = \max(a,b)$. $\endgroup$ – KCd Dec 27 '14 at 0:33
  • 1
    $\begingroup$ You ask about a place to find a proof that the minimal number of generators of $(\mathbf Z/(2))^n$ is $n$. Знаете ли Вы книгу по алгебре Винберга? $\endgroup$ – KCd Dec 27 '14 at 0:37
2
$\begingroup$

Your particular question is just "Does the vector space $\mathbf{F}_2^n$ over the field $\mathbf{F}_2$ have dimension $n$?"

For the question above that, it is not true! To get a flavor of the problem, observe that $\mathbf{Z}_2 \times \mathbf{Z}_3 \cong \mathbf{Z}_6$.

See the structure theorem for finitely generated modules over a principal ideal domain. Your question is the special case where the principal ideal domain is just the integers, so that "module" means "abelian group". "Invariant factors" and "Smith normal form" are probably things to pay close attention to.

$\endgroup$

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.