# Finitely generated abelian group isomorphic to infinite abelian group?

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?

• In the first sentence, $A$ is a finitely generated free abelian group. In the second sentence, $A$ is a finitely generated torsion abelian group. – Zev Chonoles Sep 28 '13 at 20:52
• which is a free abelian group which is torsion. – Shine On You Crazy Diamond Sep 28 '13 at 20:53
• The only abelian group that is both free and torsion is the trivial group. – Zev Chonoles Sep 28 '13 at 20:53
• So nontrivial torsion groups don't have a basis? – Shine On You Crazy Diamond Sep 28 '13 at 20:54
• @EnjoysMath Yes. The independence condition implies that a torsion element can't be part of a basis. – user61527 Sep 28 '13 at 20:56

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 $y$?)