If $|G|=pq$ and $|\langle a\rangle|=p$ then $\langle a\rangle$ is normal Let $|G|=pq$, where  $p,q$ are primes with $p>q$ and $|\langle a\rangle|=p$ for some $a\in G$ then $\langle a\rangle=A$ is normal

Since $|\langle a\rangle|=p$, A is a subgroup with order $p$, from the 3rd Sylow theorem we get that $|\text{Syl}_p(G)|\equiv1\mod{p}$.
Now if $|\text{Syl}_p(G)|=1$ I am done because I know that a Sylow-$p$-group is normal iff  $|\text{Syl}_p(G)|=1$, but if  $|\text{Syl}_p(G)|=p+1$ I don't know how I can lead up to a contradiction
 A: Since p>q, p does not divide q-1 and hence $n_p$ =1 and hence Sylow-p is normal.
A: There is a much simpler way to prove this:
Proposition Let $p$ be the smallest prime dividing the order of the finite group $G$, and assume that the subgroup $H$ has index $p$. Then $H \lhd G$.
Lemma Let $H \leq G$ with $G=HH^g$ for some $g \in G$. Then $G=H$.
Proof Apparently $g=hk^g$, for some $h,k \in H$. Hence $g=hg^{-1}kg$, from which one derives that $g=kh \in H$. Hence $H^g=H$, so $G=HH^g=HH=H$.
Let us proceed with a proof of the Proposition: assume that $H$ is not normal, so there exists a $g \in G$ with $H^g \ne H$. Since $H \ne G$, we have $|G| \gt |HH^g|$. But $|HH^g|=\frac{|H| \cdot |H^g|}{|H \cap H^g|}$. This shows that $|G:H|=p \gt |H^g:H \cap H^g|$. We conclude that $|H^g:H \cap H^g|=1$, as $p$ is the smallest prime dividing $|G|$. This means that $H^g=H \cap H^g$, so $H^g \subseteq H$, implying $H=H^g$, since $H^g$ and $H$ have the same order. We arrive at a contradiction.
A: This argument uses solely the class equation.
In general, if $|G|=pq$, with $p$ and $q$ distinct primes, then $G$ has center either trivial or equal to the whole group (see here, again with a bare class equation argument). If $Z(G)=G$, then $G$ is Abelian and you are done.
If $Z(G)$ is trivial, then the class equation reads:
$$pq=1+k_pp+k_qq \tag 1$$
where $k_i$ is the number of the conjugacy classes of size $i$.  Now,  there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$)$^\dagger$. Since each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$.

*

*Case $q\nmid p-1$. Then $q\mid m$, and hence $(1)$ yields:
$$pq=1+k_pp+m'q(p-1) \tag 2$$
for some positive integer $m'$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ for some positive integer $n$, which replaced in $(2)$ yields:
$$p=n+m'(p-1)$$
In order for $m'$ to be a positive integer, it must be $n=1$ (which in turn implies $m'=1$). So, $1+k_pp=q$: but this is a contradiction, because $p\nmid q-1$ for $p>q$. So $G$ must be Abelian, and you are done.

*Case $q\mid p-1$. Likewise as before, there are exactly $k_pp$ elements of order $q$ (they are the ones in the conjugacy classes of size $p$)$^\dagger$. Since each subgroup of order $q$ contributes $q-1$ elements of order $q$, and two subgroups of order $q$ intersect trivially, then $k_pp=l(q-1)$ for some positive integer $l$. Then $(1)$ yields:
$$pq=1+k_qq+l(q-1)$$
whence:
$$l=\frac{(p-k_q)q-1}{q-1}$$
In order for $l$ to be a positive integer, either $k_q=p-1$ or $k_q=\frac{p-1}{q}$. In the former case, $(1)$ reads:
$$pq=1+k_pp+q(p-1)$$
whence:
$$0=1+k_pp-q$$
and finally:
$$k_p=\frac{q-1}{p}$$
contradiction, because $p>q$ and then $k_p$ is not an integer. So we are left with $k_qq=p-1$, i.e. there is one subgroup of order $p$, only, which is then normal.


$^\dagger$An element lies in a conjugacy class of size $q$ (respectively, $p$) if and only if its order is $p$ (respectively, $q$). This claim follows from the orbit-stabilizer theorem and the fact that, for $g\in G\setminus\{e\}$, it is $\langle g\rangle=C_G(g)$.
