I don't understand a part of this lemma.
Lemma. 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 $p$.
Proof. We first prove by induction that if $G$ has exponent $n$ then the order of $G$ divides some power of $n$. Let $b\in G$, $b\neq 1$, and let $H$ be the cyclic subgroup generated by $b$. Then the order of $H$ divides $n$ since $b^n=1$, and $n$ is an exponent for $G/H$. Hence the order of $G/H$ divides a power of $n$ by induction, and consequently so does the order of $G$ because $$ (G:1)=(G:H)(H:1). $$ Let $G$ have order divisible by $p$. By what we have just seen, there exists an element $x$ in $G$ whose period is divisible by $p$.
Where is there any hint in the above that there is such an $x$?