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 $G.$ (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).
 A: By David Wheeler's comment, your proof for 1) is correct. I offer alternative proof below if you are interested.
Using Sylow's theorems we can arrive at the result. For starters, in any group of order $35$ there is a subgroup of order $7$ in $G$ by Sylow's first theorem. By Sylow's third theorem we have that the number of Sylow subgroups of order $7$,say $x$ for simplicity, is congruent to $1$ modulo $7$. So $x=1+7k$ for some $k$. But $x$ also divides $[G:H]$ by Sylow's third theorem. Thus $x|5$ leaving that there is only $1$ subgroup of order $7$. Finally, we conclude that $H$ is normal by Sylow's second theorem, which states that all Sylow $p$ subgroups are conjugate to one another. Since there is only one subgroup of order $7$, we have $gHg^{-1}=H$ for every $g\in G$.
As for 2, I suggest doing the following:
Suppose $|G|=35$ and that $G$ is an abelian group. By Cauchy's theorem for finite groups we have that $G$ contains subgroups of order $5$ and $7$. Say $o(x)=5$ and $o(y)=7$. Since $G$ is abelian, we have $o(xy)=\operatorname{lcm}(x,y)=35$. As $G$ contains an element of order $35$, $xy$, we have that $G$ is cyclic, generated by $xy$.
