If G is a group of order n=35, then it is cyclic 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. 

Here's a counting argument.  If $G$ is not cyclic, then every element of $G$ is of order $1$, $5$, or $7$, and, as noted above, no element of order $5$ commutes with any element of order $7$.
Let $G_5$ be the elements of order $5$.  We can see that $4\mid|G_5|$ by partioning $G_5$ into sets $\{g,g^2,g^3,g^4\}$.
On the other hand, given a $y$ of order $7$, we can partition $G_5$ into sets of $7$ elements, since we can separate $G_5$ into sets $\{g,ygy^{-1},y^2gy^{-2},...,y^6gy^{-6}\}$  (Note that, if $y^igy^{-i} = y^jgy^{-j}$, then $y^{i-j}g = gy^{i-j}$.  So this must give $7$ distinct values, or some $y^{i-j}$ commutes with $g$, which would imply that $G$ is cyclic.)
So, $28\mid |G_5|$.  Similarly, $30\mid |G_7|$.  But $35 = 1 + |G_5| + |G_7|$

Note, you don't actually need Sylow to show that there are elements of order $5$ and $7$ here, because if $G$ is not cyclic, you have that $35=1+|G_5|+|G_7|$ and $|G_5|$ is a multiple of $4$ and $|G_7|$ is  multiple of $6$, so it is not possible for either to be $0$.
