Show that $\mathbb{Z}_2$ is not a free $\mathbb{Z}_6$-module

Question

Let $ \Lambda = \mathbb{Z}_6 $，the ring of integers modulo $ 6 $.Since $ \mathbb{Z}_6 = \mathbb{Z}_2 \oplus \mathbb{Z}_3 $ as a $\mathbb{Z}_6 $-module,then $\mathbb{Z}_2$ as well as $\mathbb{Z}_3$ are projective $\mathbb{Z}_6$-modules.How ,they are plainly not free $\mathbb{Z}_6$-modules.

I know that if each $P_{i}$ is projective then ${ \bigoplus }_{ i\in I }{ P }_{ i }$ is projective and the converse is true,so what I cannot know is why they are not free $\mathbb{Z}_6$-modules.

Thanks in advance.

Answer 1

Hint: A finite free $\mathbb{Z}_6$-module has order a multiple of $6$.