# How can I find the Sylow $p$-subgroup which is not commutative?

Let the Sylow $$p$$-subgroup $$P$$ of the group, $$G$$

Then can we say the Sylow $$p$$-subgroup always commutative?

This is definitely true when the order of $$P$$ is a prime number or its square. But the other case, like the order is a cubic or a higher power of the prime number, $$p$$.

I couldn't find any counterexample.

A popular counterexample at this site are the $$2$$-Sylow subgroups of $$S_4$$:
Are all Sylow 2-subgroups in $S_4$ isomorphic to $D_4$?
The dihedral group $$D_4$$ of order $$8$$ is clearly non-abelian.
There are many groups of prime power order which are not abelian. Take one of the non-abelian groups of order $$8$$ and make the direct product with a group of order $$3$$. You have a group of order $$24$$ which has a non-commutative Sylow $$2$$-subgroup (by construction).
• I was going to put $S_4$ in my answer, since it is the most familiar example, but I think it is also worth drawing attention to the fact that we can sometimes build groups with the properties we want using quite simple techniques - and if we build them in, they are trivial to prove. – Mark Bennet Jun 30 '19 at 14:59