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I was trying the following: If F is a free abelian group finitely generated by n elements then rank of F is less than or equal to n.

I think we cannot proceed like in vector spaces because here the coefficients are integers and not coming from a field. Then how to do?

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    $\begingroup$ What is your definition of "rank of (free) abelian group"? For me it is precisely what you give: the cardinality of a minimal set of generators, or the cardinality of any set of free generators. $\endgroup$ – Timbuc Mar 15 '15 at 15:50
  • $\begingroup$ @Timbuc Sir for me it is the cardinality of the basis. $\endgroup$ – akansha Mar 15 '15 at 15:53
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    $\begingroup$ @akansha Do you have that the cardinality of any spanning set is greater than or equal to the cardinality of any basis? $\endgroup$ – Joe Johnson 126 Mar 15 '15 at 15:55
  • $\begingroup$ @JoeJohnson126 Sir like for the set of integers. $\endgroup$ – akansha Mar 15 '15 at 15:57
  • $\begingroup$ @akansha I'm not sure what you mean. What I mean is, if $\{x_1,\ldots,x_k\}$ is a spanning set of a free abelian group, then any basis has to have less than or equal to $k$ elements. $\endgroup$ – Joe Johnson 126 Mar 15 '15 at 15:59
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Write $F \cong \mathbb Z^r$. Consider the homomorphism $\pi: F \cong \mathbb Z^r \to (\mathbb Z/2\mathbb Z)^r = \mathbb F_2^r$ given by the projection $\mathbb Z\to \mathbb Z/2\mathbb Z$ on each component. If $F$ is generated by $n$ elements $x_1,\dots,x_n$, then their images $\pi(x_1),\dots,\pi(x_n)$ span the $r$-dimensional vector space $\mathbb F_2^r$, hence we must have $n\geq r$.

(Obviously there is nothing special about the choice of prime 2 in this proof. Any other prime would also work.)

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