(1) Suppose $G$ is a group with $|G|=35$. Prove that if $H$ is a subgroup of $G$ with order $7$, then $H$ is a normal subgroup of $G.$
(2) Suppose that G is a group with $|G|=35.$ Prove that if $G$ is abelian, then $G$ must be a cyclic group.
Ok, so here is what I have so far. We have to suppose to that $G$ is a group and that $H$ is a subgroup of finite index. Let $n=[G:H]$. We know to be true that there exists a homomorphism $\varphi:G \to S_n$ s.t. $Ker(\varphi) \subseteq H.$
Suppose that $G$ is a group of order $35$ and that $H$ is subgroup of order $7$. Then we have, $n=[G:H]=|G|/|H|=35/7=5$ By the Cayley Homomorphism theorem, there exists $\varphi :G \to S_5$ s.t. $Ker(\varphi) \subseteq H$. Let $K=Ker(\varphi)$. Thus, $K$ is a subgroup of $H$ and by Lagrange's theorem, $|K|$ divides $|H|=7$, therefore $|K|=1$ or $7$.
If $|K|=1$, then $\varphi$ would be injective and therefore $G$ would be isomorphic to $\varphi(G),$ a subgroup of $S_5$ which means that $S_5$ would have a subgroup of order $35$ and by Lagrange, $S_5 \mid 35$ but $|S_5|=120$ and $120 \not \mid 35$. Therefore, $|K| \neq 1$.
That proves $|K|=7$ and since $K \subseteq H$ and $|H|=7,$ it follows that $H=K$ and since $K$ is the kernel of a homomorphism, $K$ must be a normal subgroup of $G$. Hence $H$ is a normal subgroup of $G$.
I am hoping that this is correct. I am not sure what else I need or how to go on to prove (2).