Non-commuting coprime elements in finite non-abelian groups Let G be a finite non-abelian group having two distinct primes dividing its order.
Is it always true that G has two non-commuting Sylow subgroups?
$\hskip300pt$ Thank you.
EDIT: I asked in a comment if the answer is also no if G is non-nilpotent. The answer turns out now to be affirmative.
 A: The answer to the question as it stands, is no. A non-Abelian nilpotent group $G$ of order $p^{a}q^{b}$ where $p,q$ are distinct primes and $a,b$ are positive integers has
only two Sylow subgroups (one for $p$ and one for $q$) say $P$ and $Q$ respectively.
These commute with each other in the strongest sense that every element of $P$ commutes with every element of $Q$.
Perhaps a more natural condition for subgroups is whether they are permutable.
Two subgroups $A$ and $B$ are said to be permutable if $AB = BA$. This does not necessarily mean that all elements of $A$ commute with all elements of $B$.
In a finite nilpotent group, any two Sylow subgroups are permutable, and this condition 
characterizes nilpotent groups (I do not know who was the first to observe this). If $G$ is a finite group which is not nilpotent, $G$ has a Sylow $p$-subgroup $A$ which is not normal for some prime $p.$ Then $G$ must have another Sylow $p$-subgroup $B.$ The set $AB$ has cardinality a power of $p$ greater than $|A|$, so $AB$ is not a a subgroup of $G$ by Lagrange's theorem, so it is not the case that $AB = BA$, and $A$ and $B$ are not permutable.
Later edit: In response to an expansion of the original question, it is possible to go further. If $G$ is a finite group which is not nilpotent, then there are two primes $p$ and $q$ and an element $x \in G$ of order a power of $p$ and an element $y \in G$ of order a power of $q$ such that $x$ and $y$ do not commute. There may be a simpler way to see this, but one way to do it is to use Frobenius's normal $p$-complement theorem.
Since $G$ is not nilpotent, there is a prime $p$ such that $G$ has no normal $p$-complement. By the mentioned theorem of Frobenius, there is a $p$-subgroup $P$ of $G$ (which need not be Sylow) such that $N_{G}(P)/C_{G}(P)$ is not a $p$-group. For some prime $q \neq p,$ there is then an element $y$ of $q$-power order in $N_{G}(P) \backslash C_{G}(P)$. There must be an element $x \in P$ such that $xy \neq yx,$ and we have the desired two elements.
A: I don't know that you mean by "non-commuting Sylow subgroups", but whatever you mean, the answer is no. If you mean subgroups $\text{Syl}_p$ and $\text{Syl}_q$ such that there exist $g\in\text{Syl}_p$ and $h\in\text{Syl}_q$ that don't commute, then any direct product $\text{Syl}_p\times \text{Syl}_q$ of a non-commutative $p$-group by any $q$-group is a counterexample. If you mean subgroups as above such that $\text{Syl}_p\text{Syl}_q\neq \text{Syl}_q\text{Syl}_p$, then the same counterexample works. Finally, if instead of "non-commuting" you meant non-commutative, then any non-abelian semidirect product $C_p\rtimes C_q$ is a counterexample.
