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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.

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A popular counterexample at this site are the $2$-Sylow subgroups of $S_4$:

Sylow 2 subgroups of S4

Are all Sylow 2-subgroups in $S_4$ isomorphic to $D_4$?

The dihedral group $D_4$ of order $8$ is clearly non-abelian.

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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).

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  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Jun 30 '19 at 14:59

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