Let $G$ be a finite abelian group and let $p$ be a prime that divides order of $G$. then $G$ has an element of order $p$
Proof When $G$ is abelian. First note that if $|G|$ is prime, then $G \approx Z_p$ and we are done.
In general, we work by induction. If $G$ has no nontrivial proper subgroups, it must be a prime cyclic group, the case we’ve already handled.
So we can suppose there is a nontrivial subgroup $H \leq G$. Either $p$ divides $|H|$ or $p$ divides $|G/H|$.
In the first case, by induction, $H$ has an element of order $p$ which is also of order $p$ in $G$ so we’re done.
In the second case, if $g H$ has order $p$ in $G/H$ then $|g H|$ divides $|g|$, so $\langle g \rangle \approx Z_{kp}$ for some $k$, and $kg$ ∈ G has order $p$.
I am confused over how
(i) Either $p$ can divide $|H|?$
(ii) In the second case, if $g H$ has order $p$ in $G/H$ then $|g H|$ divides $|g|$, so $\langle g \rangle \approx Z_{kp}$ for some $k$, and $kg$ ∈ G has order $p$
$p$ was supposed to divide $|G|$ . but why does the statement say that $p |~~ ||H|$
Help will be really appreciated. Thanks
For the alternative approach that @user133281 suggested, see Isaacs' 'Finite Group Theory' exercise 1A.8. Using group actions, it is easier, I think.