Prove that a group is cyclic $G$ is abelian of order $35$. and for all $x\in G$, $x^{35}=e$. I need to show that $G$ is cyclic. This seems perfectly obvious but I dont know how to write the proof. Help would be appreciated!
Thanks in advance!
 A: In general if an abelian group $G$ is of order $pq$ with $p$ and $q$ being different primes then $G$ is the cyclic group of order $pq$.
In this case, take $x\in G$ such that $x\ne e$. Let $\langle x\rangle$ be the subgroup of $G$ generated by $x$. If $|\langle x\rangle|=35$ then $G=\langle x\rangle$ so is cyclic. If not, then $|\langle x\rangle|$ divides $|G|$ so is $5$ or $7$. The two cases can be treated in essentially the same way, so I will take $|\langle x\rangle|=5$ and $|\langle x\rangle|=7$ can be treated similarly. 
With $|\langle x \rangle|=5$, let $y\in G\setminus\langle x\rangle$, then $y\langle x\rangle$ is a non-identity element of $G/\langle x\rangle$. $|G/\langle x\rangle|=7$ so $y^7\in\langle x\rangle$. If $y^7\ne e$, then we must have $|y|=35$. If $y^7=e$ then $|yx|=35$. In either case we've shown that $G$ is cyclic.
A: By a theorem of Cauchy, $G$ has an element $x$ of order $5$ and an element $y$ of order $7$. Since $G$ is abelian, $x$ and $y$ commute, and hence the order of $xy$ is the l.c.m. of $5$ and $7$, which is $35$. So $G=\langle xy \rangle$ is cyclic.
A: You don't need the hypothesis that $G$ is abelian: if $G$ is a group of order $pq$ with $p<q$ and $p$ doesn't divide $q-1$ then  $G$ is cyclic of order $pq$, in particular abelian. 
Proof By Cauchy there exist elements $a,b$ of order $p$ and $q$ respectively. Then $H=\langle a\rangle$ and $K=\langle b\rangle$ are Sylow subgroups of $G$ of said orders. Since $[G:K]=p$ is the least prime that divides $|G|$, $K$ is normal in $G$. Since $H$ and $K$ have coprime orders, it follows that $H$ and $K$ have trivial intersection. Thus $G=K\rtimes H$, and it suffices to show this semidirect product is in fact direct, i.e. the action of $H$ by conjugation on $K$ is the identity. Now ${\rm Aut}(K)$ has order $q-1$, consider the map $\varphi: H\to {\rm Aut}(K)$ that sends the generator $a$ to the automorphism of $K$ that sends $b\mapsto aba^{-1}$. Since $a$ has order $p$, the image of $\varphi(a)$ must have order a divisor of $p$. Since $\varphi(a)$ is an element of a group of order $q-1$, its order divides $q-1$. But we know that $p$ doesn't divide $q-1$, so that the order of $\varphi(a)$ is a divisor of $(p,q-1)=1$, i.e. $\varphi(a)$ is the identity autmorphism of $K$, and $aba^{-1}=b$, so $a,b$ commute and $G$ is abelian cyclic of order $pq$.
A: any element except the identity must have order $5,7$ or $35$.
suppose there were no element of order $35$
let there be $m$ elements of order $5$ and $n$ elements of order $7$
if $mn \gt 0$ we may choose two elements $a$ of order $5$ and $b$ of order $7$
but then $a+b$ has order $35$, a contradiction
so $mn=0$ and either $m=34, n=0$ or $m=0,n=34$. take the former case.
every element belongs to a subgroup of order $5$. 
any two such subgroups are either identical or overlap in the identity alone. 
thus if the number of such subgroups is $M$, we must have
$$
4M = 34
$$
but M is an integer, therefore we have a contradiction.
the other case may be treated in the same way since $6$ does not divide $34$
hence there must be an element of order $35$
