I have to prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective.
I've found already this question Prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective and injective. Find a $\mathbb{Z}/4\mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).
My idea is to use the fact that $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \bigoplus \mathbb{Z}/3\mathbb{Z}$, these two submodules are projective and they are fields, hence all $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$-modules are free. But I don't know if taken any $\mathbb{Z}/6\mathbb{Z}$-module I can "decompose" it in a sum of a $\mathbb{Z}/2\mathbb{Z}$-module and a $\mathbb{Z}/3\mathbb{Z}$-module.
Best Answer
Let $M$ be a $\Bbb{Z}/6\Bbb{Z}$ module. We'll show that $M=2M\oplus 3M$.
If $m\in 2M\cap 3M$, then $m=2m'=3m''$ for some $m',m''\in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$ so $2M\cap 3M=0$.
On the other hand, $m=3m-2m$, so $2M+3M=M$.
Thus $M=2M\oplus 3M$. Note that $2M$ is a $\newcommand\ZZ{\Bbb{Z}}\ZZ/3\ZZ$ module and $3M$ is a $\ZZ/2\ZZ$ module.
Thus the claim you wanted to prove is true.
Then $2M$ is a direct sum of copies of $\ZZ/3\ZZ$ and $3M$ is a direct sum of copies of $\ZZ/2\ZZ$, and as you've noted $\ZZ/3\ZZ$ and $\ZZ/2\ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.