$12 = 2^2\times 3$. The number of non-isomorphic abelian groups of order $12$ is the product of the number of partitions of each prime power: $p(2^2)=2$ (partitions of 2: $2$ and $1+1$, giving $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$) times $p(3)=1$. Total: $2\times 1 = \mathbf{2}$.