I'm trying to solve this question in page 74 of Hungerford's book:

A free abelian group is a free group (Section I.9) if and only if it is cyclic.

I have no idea how to proceed, a solution or a hint would be welcome.

This question helped me to see the difference between free abelian groups and free groups, but I still can't solve the problem.

Thanks in advance


Following the comments, my attempt of solution is:


If $F$ a free abelian group is a free group itself, then it generated by one element, then it's cyclic.


Using the universal property:

Let $F$ be the free group which is cyclic on a set $X$ and $i:X\to F$ the inclusion map. If $G$ is a group and $f:X\to G$ a map of sets, then define $\bar f:F\to G$, $\bar fi(x)=f(x)$, this $\bar f$ is well-defined and unique by definition. (I didn't use the fact $F$ is cyclic)

I would like to know if I am right.

Thanks again

  • 4
    $\begingroup$ A free group generated by more than one letter is non-abelian; different generators don't commute. $\endgroup$ – user61527 Jun 4 '14 at 3:19
  • 1
    $\begingroup$ Hint: free groups satisfy a universal property. Any map from a generating set to any group can be extended to a homomorphism of the free group. If the image is nonabelian, then the free group must also be nonabelian. $\endgroup$ – Cheerful Parsnip Jun 4 '14 at 3:35

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