Suppose $G$ is an non-commutative group of some finite order say $n$.
Find all $n$ where $n$ is some possible order of a Non-Commutative Group $G$ where it is always possible to find two elements $a,b\in G$ such that $\gcd(o(a),o(b))$ is composite?
My try:
I checked the groups $D_4,Q_8$.
I found that in $D_4$ we can find $r,r^2$ such that $o(r)=o(r^3)=4$.
Also the result holds in case of $Q_8$ as we can find the matrices
$\begin{bmatrix} 0 & -i\\ i &0\end{bmatrix}$ and $\begin{bmatrix} 0 & -i\\ i &0\end{bmatrix}$ which have orders $4$.
Also I found from previous question I posted here that the result fails for non-abelian groups of order $pq,2^n$ where $q\equiv 1 \mod p$
Related :Is it always possible to find two elements $a,b\in G$ such that $\gcd(o(a),o(b))$ is composite?
How to find all such $n$ in general
Any help will be highly appreciated.
If $G$ has no element of composite order, then there are clearly no such elements $a,b$.
If $G$ has an element $g$ of composite order, then $g^{-1}$ is a different element of the same composite order, so you can use $g$ and $g^{-1}$.
So the question is, for which values of $n$ does every nonabelian group of order $n$ have an element of composite order?
I don't know.
Certainly it's true for $n=2^m$, $m\ge3$. But I don't even know whether every nonabelian group of order $36$ has an element of composite order. (Although it shouldn't be too hard to look this up somewhere, as there are tables of small groups on the web.)