TITLE: Abelian subgroup in an infinite non-abelian $3$-group
QUESTION [6 upvotes]: Does an infinite non-abelian $3$-group has an infinite abelian subgroup? 

This result holds for $2$-groups, but I wonder whether this holds or not for $3$-groups. 
Thanks in advance.

REPLY [4 votes]: The answer is affirmative if $G$ has exponent $3$. 
Recall that a group $G$ is locally finite if every finite subset of $G$ generates a finite subgroup, that is if every finitely generated subgroup of $G$ is finite.  It follows from a well known result on Burnside Problem, that every 3-group is locally finite.
Now from a result of P. Hall, Kulatilaka and Kargapolov (which can be found in Robinson's book
"A course in the theory of groups", Section 14.3), one knows that every locally finite group which is infinite, contains an infinite abelian subgroup.  
As, I think you have not imposed a condition on the exponent of $G$, I think there is $2$-generators groups of exponent $3^n$ with $n$ sufficiently large, in which every proper subgroup is cyclic, so do not contain infinite proper subgroup