Isn't every finite cyclic group a free group?

Question

I thought that, as a finitely generetade group, every cyclic group was free. But in homological algebra we have the result "If $F$ is a free group then $H_i(F,A)=H^i(F,A)=0$ for every $i\geq 2$ and for every left $F$-module $A$." Though we also have the following result "If G is a finite cyclic group of order $n$, then
$$H_i(G,\mathbb{Z})=\left\{\begin{array}{l}\mathbb{Z},\textrm{ if }i=0\\ \mathbb{Z}_n,\textrm{ if }i\textrm{ is odd}\\ 0,\textrm{ if $i$ is even}\end{array}\right."$$ How is that not contraditory?

Answer 1

Recall, that any nontrivial finite group cannot be free, since the elements of a free generating set of a free group have infinite order. The infinite cyclic group is free, of course.