Lang's Algebra says that if an abelian group $A$ is free and finitely generated by $(x_i), i=1,\dots, n$ , then it is isomorphic to $\mathbb{Z}x_1 \bigoplus \cdots \bigoplus \mathbb{Z}x_n$, which is isomorphic to $\mathbb{Z}\bigoplus\cdots\bigoplus\mathbb{Z}$ ($n$-fold). But if $A = A_{\text{tor}}$, i.e. a torsion group, then isn't $A \approx \mathbb{Z}/m_i\mathbb{Z}\bigoplus\cdots\bigoplus\mathbb{Z}/m_n\mathbb{Z}$, where $m_i = $ exponent of $x_i$, finite? So can we have a finite group isomorphic to an infinite one?
$\begingroup$
$\endgroup$
7
asked Sep 28 '13 at 20:50
-
$\begingroup$ In the first sentence, $A$ is a finitely generated free abelian group. In the second sentence, $A$ is a finitely generated torsion abelian group. $\endgroup$ – Zev Chonoles Sep 28 '13 at 20:52
-
$\begingroup$ which is a free abelian group which is torsion. $\endgroup$ – StudySmarterNotHarder Sep 28 '13 at 20:53
-
$\begingroup$ The only abelian group that is both free and torsion is the trivial group. $\endgroup$ – Zev Chonoles Sep 28 '13 at 20:53
-
$\begingroup$ So nontrivial torsion groups don't have a basis? $\endgroup$ – StudySmarterNotHarder Sep 28 '13 at 20:54
-
$\begingroup$ @EnjoysMath Yes. The independence condition implies that a torsion element can't be part of a basis. $\endgroup$ – user61527 Sep 28 '13 at 20:56
|
Show 2 more comments
$\begingroup$
$\endgroup$
No, a finite group and an infinite group can't be isomorphic. $\mathbb Z$ (or $\mathbb Z\cdot x$) is not isomorphic to $\mathbb Z/m\mathbb Z$.
If one group has an element $y$ of infinite order, and a second group has no such element, then it's easy to show that the groups are not isomorphic. (Where would the isomorphism map $y$?)
answered Sep 28 '13 at 20:53
