Is this true for about abelian generating sets?

Question

Given a finite abelian group $G$ which is genertaed by a set $S$.

Question : Is there always exists a $S' \subseteq S$ which is a basis of $G$?

Seems true to me for example $\mathbb{Z}_6$, take $S = \{1,2,3\}$ contains a subset $\{2,3\}$ which is a basis.

A set $B$ is said to be a basis of a abelian group $G$ if $G=\langle B \rangle$.

Answer 1

No: in $\mathbb{Z}_{12}$, $S=\{2,3\}$ generates $\mathbb{Z}_{12}$, but $\{2\}$ and $\{3\}$ don't, and $\mathbb{Z}_{12} \neq \langle 2 \rangle \times \langle 3 \rangle$.