Let $G$ be an additive cyclic group of order $pn$ where $p$ is prime. Let $a$ be any random element of $G$. I want to prove that scalar multiplying the element $a$ by $n$ (same as adding it $n$ times) gives me an element which belongs to the prime order subgroup of $G$ of order $p$.
This is what I tried.
Let $b = na$. I have to prove that $b$ belongs to the subgroup of order $p$
$(np)a = 0$ (because order of $G$ is $np$)
$p(na) = 0$
$pb = 0$
I don't think $pb = 0$ proves that the order of $b$ is $p$ - because there can be $m < p$ which also satisfies $m.b = 0$ in which case $b$ wouldn't be an element of order $p$.
Since $p$ is a prime, $m$ cannot be a factor of $p$. So $p = qm + r$ with $r$ smaller than $m$ & $r \ne 0$.
$p.b = 0$
So $(qm)b + rb = 0$
If $m$ is the order of $b$, then $mb = 0$, then $(qm)b = 0$. So $rb = 0$.
However, this is not possible because the order of $b$ is $m$ & with an $r < m$, $rb \ne 0$. Hence our assumption that there is an $m$ smaller than $p$ which is the order of $b$ is incorrect.
So we have established that the order of $b$ is $p$. Any element of order $p$ belongs to the subgroup of order $p$. Hence proved.
Is my proof correct? If it is indeed correct, is it the simplest proof or are there any simpler ways this can be proved?
I could do one touch up, but I think your solution is otherwise correct.
That's you forgot the possibility that $b=0$, or $\lvert b\rvert=1$ (but you're ok because then $b$ is in the subgroup of prime order).
At any rate, once you have $pb=0$ you are essentially done, because of the general fact: $kb=0\implies \lvert b\rvert \mid k$.
Proof:
Write $k=q\lvert b\rvert +r$, where $r\lt\lvert b\rvert $. Then $0=kb=q\lvert b\rvert b+rb=0+rb=rb\implies r=0$, by minimality of $\lvert b\rvert $.