$p^nm$ group, element of order $m$ Let $p$ be an abelian group of order $p^nm$, $p$ prime, and $p$ does not divide $m$.
Is it true that the group must contain an element of order $m$, or a multiple of $m$? If yes, how to prove it? If not, is it true in the particular case $m=p-1$?

No, not even for abelian groups with that special condition does this always holds. Let $p=5$ and $m=4$ as you request this has the form $m=p-1$, then
$$G=\Bbb Z/2\oplus\Bbb Z/2\oplus\Bbb Z/5$$
is a counter-example to both the general and specific cases.
This doesn't have an element of order $4$. If $x$ were such an element then note that $4x=x+x+x+x$ would be $0$ in the group, i.e. the identity element.
But then note that for $x=(a,b,c)$ in our direct sum, for it to be $0$ all three components must be $0$, the third component cannot have $4c=0$ unless $c=0$ since $4c\equiv-c\mod 5$ and $-c\equiv 0\iff c\equiv 0$, so if $4x=0$ then $c=0$
Then $x=(a,b,0)$. But then $a,b\in\{0,1\}$, no matter which case we have $2a\equiv 2b\equiv 0\mod 2$, so that the order divides $2$. Since $4$ does not divide $2$, we have proven the result.
