TITLE: If G is a group of order n=35, then it is cyclic
QUESTION [20 upvotes]: I've been asked to prove this. 
In class we proved this when $n=15$, but our approached seemed unnecessarily complicated to me. We invoked Sylow's theorems, normalizers, etc. I've looked online and found other examples of this approach.  
I wonder if it is actually unnecessary, or if there is something wrong with the following proof: 
If $|G|=35=5\cdot7$ , then by Cauchy's theorem, there exist $x,y \in G$ such that $o(x)=5$, $o(y)=7$. The order of the product $xy$ is then $\text{lcm}(5,7)=35$. Since we've found an element of $G$ of order 35, we conclude that $G$ is cyclic. 
Thanks.

REPLY [2 votes]: What about this one? By Sylow, there is precisely one 5-sylow subgroup and precisely one 7-sylow subgroup. There can be no other subgroup of order 5 or 7, since they would be other 5-sylow or 7-sylows and we had only one of each. Pick x in the complement to the union of the 5-sylow and 7-sylow. There are 35-5-7+1=24 elements to choose from. The order of x cannot be 1, 5 or 7. By Lagrange it divides 35 so it must be 35 and hence x generates the group so it is cyclic.