Problem
Let $G$ be a finite abelian group and $p$ a positive prime that divides $|G|$. Show that the number of elements of order $p$ in $G$ is coprime with $p$.
Let $|G|=p^nm$ with $n \geq 1$ and $gcd(p,m)=1$. If we consider $G$ as a $\mathbb Z$-module, then by the structure theorem we have $$G \cong \mathbb Z_{p^{n_1}} \oplus ... \oplus \mathbb Z_{p^{n_k}} \oplus M$$ with $\mathbb Z_{p^j} \not \subset M$ for all $j \in \mathbb N$ and with $n_1+...+n_k=n$.
I don't know what to do, I got stuck here, I would appreciate some help with this problem. Thanks in advance.
As pointed out in the comments, the claim holds in general. The $N_p$ elements of order $p$ are grouped into $\ell_p$ trivially intersecting subgroups of order $p$: $$N_p=\ell_p(p-1)\equiv -\ell_p\pmod p$$ But, as a corollary of Wielandt's proof of Sylow's theorem, $\ell_p\equiv 1\pmod p$. Therefore $$N_p\equiv -1\pmod p$$ and hence $\gcd(N_p,p)=1$.