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
Best Answer
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.