Let $\{x_1,\dots,x_r\}$ be an independent subset of a $p$-primary abelian group $G$. If $pz_i=x_i$ , then $\{z_1,\dots,z_r\}$ is independent.

Question

A set $\{x_1,\dots,x_r\}$ of nonzero elements in an abelian group is independent if, whenver there are integers $m_1,\dots,m_r$ with $\sum_{i=1}^r m_ix_i=0$, then each $m_ix_i=0$.

Let $\{x_1,\dots,x_r\}$ be an independent subset of a $p$-primary abelian group $G$. If $\{z_1,\dots,z_r\}\subset G$, where $pz_i=x_i$ for all $i$, then $\{z_1,\dots,z_r\}$ is independent.

Suppose $\sum k_iz_i=0$. Clearly, $\sum pk_iz_i=0$.
Hence $\sum k_ix_i=0$ and by the assumption $k_ix_i=0$.
Substituting $pz_i=x_i$ we get $pk_iz_i=0$.

For simplicity, here I write $pkz=0$.
Since $z$ and $x$ are nonzero, assume $|z|=p^m$ where $m\geq2$.
If $p^m \mid k$, we are done. Suppose $p^m\not\mid k$. Note that $p^m\mid pk$.
This means that $p^{m-1}\mid k$. Write $k=lp^{m-1}$ where $(l,p)=1$.
So far I can't get any contradiction to show that $l$ is divisible by $p$.

Answer 1

Good job so far. You reached the conclusion that all the $k_i$s are multiples of $p$ (because $m\ge2$). Now step back and rethink about the relation $$ 0=\sum_i k_iz_i=\sum_i (k_i/p)x_i. $$ What can you conclude now?