G5 - cauchy's theorem for abelian groups

cauchy's theorem

let $G$ be a finite abelian group, and $p$ be a prime that divides the order of $G$ , then $G$ has an element of order $p$

firstly, find an element of some prime order
consider a non-identity element $a \in G$ with $| a | = m = q n$ , where $q$ is a prime divisor of $m$
$a^{m} = a^{q n} = (a^{n})^{q} e$ , so $| a^{n} | = q$ as $a^{n} \neq e$
let $x = a^{n}$ , so $| x | = q$
as $x$ already has order $p$ , considering the case when $q \neq p$
let $H = ⟨ x ⟩$ , and $| x | = q ⟹ | H | = q$
$G$ is abelian, so every subgroup is normal, and $\bar{G} = G / H$ is a well-defined, abelian group
by lagrange's theorem:

| \bar{G} | = \frac{| G |}{| H |} = \frac{| G |}{q}

remember that $p ∣ G$ and $p \neq q$ , so $p ∣ \bar{G}$ , and $| \bar{G} | < | G |$
the statement is clearly true for $| G | = 2$
$G$ certainly has elements of prime order for if $| x | = m$ , and $m = q n$ , where $q$ is prime, then $| x^{n} | = q$
let $x \in G$ such that $| x | = q$ , for prime $q$
it is trivial for $p = q$ , so assume $q \neq p$
it is known that every subgroup of an abelian group is normal
construct $\bar{G} = G / ⟨ x ⟩$ , then $\bar{G}$ is abelian and $p$ divides $| \bar{G} |$ since $| \bar{G} | = | G | / q$