Confusion with Lang's proof of Sylow Theorem I am currently working through Lang's Algebra.  I am rather confused by what seems to be a trivial point.  In a lemma preceding the proof of the Sylow Theorem (which is essentially Cauchy's Theorem), lemma 6.1, he proves that if a finite abelian group has an exponent $n$ then its order divides some power of $n$.  I am comfortable with this fact.  However, he immediately used this to show that all such groups with order $np$, for $p$ prime, have an element of period $p$.  This seems very much like a non-sequitur to me.  What am I missing?
 A: Let’s see. If there’s no element of order $p$, mustn’t the exponent be prime to $p$? Granting that, let the exponent be $m$, but then $np | m^r$, looks like a contradiction. 
A: By the fact that $G$ has order divisible by $p$, we can assume that
$$
   |G|=p^ra
$$
where both $r$ and $a$ are positive integer and that $\text{gcd}(p,a)=1$.
Now we want to clarify the second part of this lemma using what we got in the first part. Let the exponent be $n$. With what we got in part 1, there is a positive integer $s$ such that
$$
p^ra|n^s.
$$
It can be verified that $p|n$. So for some $x \in G$ and $x \neq e$, put 
\[
 y=x^{\frac{n}{p}}\in{G}
\]
So $y^p=e$, which showed that there exists an element whose period is divisible by $p$ in $G$ as long as $p|(G:1)$.

The fact that $p^ra|n^s$ is where part 1 came into play. Say, $n^s=qp^r$, where $s,q,r$ are all positive integers. Since we have $qp^{r-1}=n^{s-1}\frac{n}{p}$ is an integer. It's immediately verified by assuming $p \nmid n$ which leads to a contradiction.
