Is $\mathbb{Z_5}$a free abelian group ? Yes/No

Question

Is $\mathbb{Z_5}$ a free abelian group ?

My attempt: I think $\mathbb{Z_5}$ is free abelian group

By the definition of free abelian group

$X$ generates $G$, and $n_1x_1 +n_2x_2 +\dots+n_rx_r=0$ for $ n_i \in \mathbb{Z} $ and distinct $ x_i \in X$ if and only if $ n_1=n_2 =\dots=n_r=0$

Take $G= \mathbb{Z_5}$ and $X=\{0,1,2,3,4\}=\{x_1,x_2,x_3,x_4,x_5\}$

Then $n_1x_1+ n_2x_2 +n_3x_3 +n_4x_4 +n_5x_5=0\implies0.n_1+1.n_2+2.n_3+3n_4+4n_5=0$

$ n_1=n_2 =\dots=n_5=0$

This satisfied the definition of free abelian group

Answer 1

Hint: Let

$$n_1=n_2 =\dots=n_5=5.$$

Answer 2

It is not a free Abelian group. A free Abelian group is either trivial or infinite.

Namely, take $a\in G, a\ne 0$ and look for $a, 2a, 3a, 4a\ldots$. Eventually (due to finiteness), two of those must be the same, say $ma=na$. Then $(m-n)a=0$ with $m-n\ne 0$. So any single non-zero element $a$ is "linearly dependent" all by itself in a finite Abelian group.