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. 

Another explicit example: 
Consider 
$$
A = \left( \begin{array}{cc}
1 & -1 
\\
0 & -1
\end{array} \right), \quad \text{and} \quad B = \left(\begin{array}{cc}
1 & 0 
\\
0 & -1
\end{array}
 \right).
$$
Then, $A^2 = B^2 = I$, but
$$
AB = \left( \begin{array}{cc}
1 & 1
\\
0 & 1
\end{array}
\right)
$$
has infinite order.
It should also be mentioned that if $x$ has order $n$ and $y$ has order $m$, and $x$ and $y$ commute: $xy = yx$, then the order of $xy$ divides $\text{lcm}(m,n)$, though the order of $xy$ is not $\text{lcm}(m,n)$ in general.  For example, if an element $g \in G$ has order $n$, then $g^{-1}$ also has order $n$, but $g g^{-1}$ has order $1$.  Joriki's example also provides a scenario where the order of $xy$ is not $\text{lcm}(m,n)$ in general.
