Lang Lemma 6.1 (before Sylow): if $p$ divides order of finite abelian group, then subgroup with $p$ order exists. Why is $x^s\neq1$ guaranteed?

Lemma 6.1. Let $G$ be a finite abelian group of order $m$, let $p$ be a prime number dividing $m$. Then $G$ has a subgroup of order $p$.

It is clear to me that $|G|$ divides power of $n$ where $n$ is exponent of a group (this is by induction, take cyclic subgroup and use Lagrange theorem). It also clear that then $p \mid n$, so we can write $n=ps$.

However, I don't understand why we have an $x$ whose period is divisible by $p$ (smallest exponent). Of course we can put $y=x^s$ so we have $y$ with period $p$, but how do we know that $x^s\neq 1$?

My idea was to take the smallest exponent (period) of a group (every finite group has an exponent obviously) and apply first part of the lemma. However, that still doesn't guarantee that $x^s\neq1$.

I feel like I'm missing something obvious, can someone enlighten me please?

PS: where would the argument fall if we use non-prime $m$ instead of prime $p$. Why in this case we cannot say there exists $x$ whose period is divisible by $m$?

We know that $p$ divides the exponent of $G$ (because $|G|$ divides a power of $n$ where $n$ is the exponent of $G$). Therefore there exists an element $y$ whose order is divisible by $p$. From now on let $y$ be of order $p\times n_0$. Then $x=y^{n_0}$ is of order $p$ whence $\langle x\rangle$ is the subgroup you are looking for.

Edit : Let us prove that such $y$ exists. We do it by contradiction. Assume such $y$ does not exists, then for any $g\in G$ the order of $g$ is prime with $p$ (this where you need $p$ prime) and therefore the exponent of $G$ is also prime with $p$.