Recall a group $\Gamma$ is said to have subexponential growth if lim sup$|E^{n}|^{1/n}=1$ for every finite subset $E\subset \Gamma. (E^{n}=\{g_{1}g_{2}...g_{n}: g_{i}\in E\}.)$
My question is: Can we prove all abelian groups have subexponential growth?
Consider any subset $E_m$ of your abelian group of size $m$. Then since the group is abelian, $|E_m^n|$ is no more than the number of ways you can choose $m$ non-negative integers to sum up to $n$, which is${{n+m-1} \choose {m-1}}$ which is no more than than $(n+m-1)^{m-1}$. Since the exponent is a constant, you will get $\lim_{n \to \infty}(n+m-1)^{(m-1)/n} = 1$. So an abelian group is indeed subexponential.