I have to show that $ \hat {6} \mathbb {Z}_{12}$ is not a projective $\mathbb {Z}_{12}$-module.
I consider the exact sequence $\hat {3}\mathbb {Z}_{12}\rightarrow \hat {6} \mathbb {Z}_{12}\rightarrow 0$ where the morphism is multiplication by $\hat {2} $.
I suppose that $\hat {6} \mathbb {Z}_{12}$ is projective. I take the identity from $ \hat {6} \mathbb {Z}_{12}$ to $ \hat {6} \mathbb {Z}_{12}$. From projectivity there exists $f: \hat {6} \mathbb {Z}_{12}\rightarrow \hat {3} \mathbb {Z}_{12}$ a morphism such that the identity is the multiplication composed with $f $.
Now I have to define $ f (\hat {6}) $.
$ f (\hat {6}) = \hat {3} $ or $ f (\hat {6}) =\hat {9} $.
But $ f (\hat {0}) =\hat {0} $(Contradiction).
It is right?
It's right. Here's an alternative way.
The submodules of the regular module $\mathbb{Z}_{12}$ (that is, over itself) are just the additive subgroups and there is a unique subgroup for each divisor of $12$, so the list is $1\mathbb{Z}_{12}$ (order $12$), $2\mathbb{Z}$ (order $6$), $3\mathbb{Z}_{12}$ (order $4$), $4\mathbb{Z}_{12}$ (order $3$), $6\mathbb{Z}_{12}$ (order $2$), $12\mathbb{Z}_{12}$ (order $1$). (You can use hats, if you prefer.)
The only subgroup of order $2$ is $6\mathbb{Z}_{12}$ which has zero intersection only with the subgroup $4\mathbb{Z}_{12}$ (apart from the zero subgroup). If $6\mathbb{Z}_{12}$ were projective, being a homomorphic image of the regular module it would be a direct summand, but it isn't.