Show that every group of order $35$ is abelian. How can I show that every group of order $35$ is abelian ? I know that if such a group is abelian the it's isomorphic to $\mathbb Z_{35}$ or $\mathbb Z_7\times \mathbb Z_5$. But, how can I show that any group of order 35 is either isomorphic to $\mathbb Z_{35}$ or to $\mathbb Z_7\times \mathbb Z_5$ ?
 A: SyLow implies that you have a subgroup of order $5$ $L$ and a subgroup on order $7$, $M$. The number $n_5$ of Sylow 5-groups divides 7 and  is equal to 1 mod 5 so is 1 so $L$ is normal. Let $x$ be an element of order $7$, the conjugation by $x$ induces an automorphism of $L$ of order $5$, so is trivial since the order of $x$ is 7, thus if $y$ generates $L$, $x$ commutes with $y$ and the group is the product $Z_5\times Z_7$.
A: edit: I apologize, but during writing this argument I did not notice that Tsemo Aristide had anticipated me and had given a sketch of what seems to be exactly the same argument(with some minor differences, since I did not use Sylow's theorem). I am looking forward for comments should I delete my answer or leave it as other people may find almost complete argument I have written usefull.
Suppose that there are two distinct subgroups $H_1$, $H_2$ inside $G$ of order $7$. Then $H_1\cap H_2=\{1\}$, otherwise we would have $H_1=H_2$. Using the fact that $H_1\cap H_2=\{1\}$ we can derive that $H_1H_2$ has cardinality $7^2=49$. This is a contradiction. Thus there is at most one subgroup of $G$ of order $7$.
By Cauchy's theorem there exists a subgroup $H\subseteq G$ of order $7$. By what we have said above such subgroup is unique. Obviously such subgroup is normal(even characteristic). 
Again by Cauchy's theorem there is a subgroup $K\subseteq G$ of order $5$. Since $H$ is normal we have homomorphism $\phi:K\rightarrow \mathrm{Aut}(H)$ given by:
$$\phi(k)(h)=k^{-1}hk$$
Next note that $K$ is of order $5$ and:
$$\mathrm{Aut}(H)\cong \mathrm{Aut}(\mathrm{Z}_7)=\mathrm{Z}_6$$
Thus there are no homomorphisms from $K$ to $\mathrm{Aut}(H)$ except for the trivial one. We deduce that $\phi$ is trivial. This means that:
$$k^{-1}hk=h$$
for every $k\in K$ and $h\in H$. Therefore, elements of $H$ and $K$ commute with each other. 
Since $H\cap K=\{1\}$, we derive that $G=HK$. Hence $G$ is abelian.
A: These facts may be helpful.


*

*$|G|=p×q$, where $p$ and $q$ are primes and $p<q$. If $p$ doesn't divide $q-1$ then $G$ is cyclic.

*$G$ is cyclic $\rightarrow G$ is abelian.

