2
$\begingroup$

I am trying to understand the basic theory of free abelian groups and let me ask you the following question. This is the excerpt of the book which I'm reading.

Lemma (Universal property of free abelian group): Let $X$- theset of generators of abelian group $F$, $|X|=n$. Then the following statements are equivalent:

(i) $F=F^{ab}_n$ - free abelian groups and $X$- the set of its generators.

(ii) Any mapping $\varphi:X\to A$ where $A$ - abelian group induces the homomorphism $\tilde{\varphi}:F\to A$.

Corollary: Any abelian group with $n$ generators is homomorphic image of some free abelian group $F^{ab}_n$.

Proof: Follows immediately from univeral property.

However, for me is not clear how it follows from lemma. But I have done in the following way: suppose that $(A,+)$ is abelian group with generators $\{a_1,a_2,\dots, a_n\}$. Let $F^{ab}_n$ be some free abelian group with generators $X=\{x_1,\dots,x_n\}$. Consider the mapping $\varphi:X\to A$ by $\varphi: x_i\mapsto a_i$ then we can consider the homomorphism $\tilde \varphi: F^{ab}_n\to A$ defined via $$\tilde \varphi:\sum k_ix_i \mapsto \sum k_ia_i.$$

Then easy to see that $\tilde \varphi(F^{ab}_n)=A$.

Is my proof looks correct?

But I was not able to derive it immediately from Universal property.

Would be very grateful for help!

$\endgroup$
  • 1
    $\begingroup$ The condition (ii) seems to be completely out of way. The usual (for me) universal property (ii) states that any map $\;\phi: X\to A\;$ , with $\;A\;$ any abelian group, can be uniquely extended to a homomorphism of abelian groups $\;\overline\phi: F_n^{ab}\to A\;$ ...Nothing about bijection. $\endgroup$ – DonAntonio Nov 17 '18 at 22:06
  • $\begingroup$ @DonAntonio, Is my proof looks correct? $\endgroup$ – ZFR Nov 17 '18 at 22:13
  • $\begingroup$ Yes, it is. Extremely easy...and useful. Delete that "bijective" there. Completely nonsense. $\endgroup$ – DonAntonio Nov 17 '18 at 22:15
  • $\begingroup$ @DonAntonio, thanks a lot! Firstly I thought that without bijection the induced homomorphism will not be surjective. $\endgroup$ – ZFR Nov 17 '18 at 22:17
  • $\begingroup$ If you leave the word "bijective, it'd mean $\;|A|=n\;$ , which of course not necessarily is true. It'd be true if you'd ask that the map be from $\;X\;$ to a generating set with $\;n\;$ elements of $\;A\;$ ...! Observe you also teh very important property of uniqueness ...which, btw, you did not prove as you didn't mention it. Read about this universal property anywhere in the web...even in Wikipedia $\endgroup$ – DonAntonio Nov 17 '18 at 22:25
1
$\begingroup$

You are sort of partially reproving the universal property for this case, which is not needed.

The universal property ought to be stated as

For every function $\phi: X \to A$ where $A$ is an Abelian group, there is a unique homomorphism $f:F= F^{\textrm{ab}}_n \to A$ such that $f| X=\phi$.

Now, if $A$ is Abelian group with your $n$ generators, indeed define $\phi(x_i) =a_i$ when $\{x_1,\ldots,x_n\} = X$. Then by the universal property, a homomorphism $f: F \to A$ exists, extending $\phi$. As $f[X]=\phi[X]$ is a set of generators for $A$, $f[F]=A$, as $f[F]$ is a subgroup containing the $a_i$, and the smallest subgroup containing the $a_i$ is $A$ (that is what it means to be a generating set). So surjectivity is almost automatic, and we're done proving the corollary.

$\endgroup$
  • $\begingroup$ Very nice and good answer! +1 $\endgroup$ – ZFR Nov 17 '18 at 23:59

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.