A condition for a finite group $G$ be nilpotent

Question

Is true that a finite group $G$ is nilpotent if and only if $[x,y]=1$ for all $x,y \in G$, such that $(\mid x\mid, \mid y \mid) = 1$, where $[x,y] = x^{-1}y^{-1}xy$, ie, is the commutator of $x$ and $y$; and $(a,b)$ denotes the greatest common divisor of the integers $a$ and $b$?

Answer 1

Yes, this is true. The gist is that Sylows are normal in a nilpotent group.

If $P$ is a Sylow $p$-subgroup, then $N_G( N_G(P) ) = N_G(P)$ (abnormality). However, in a nilpotent group $N_G(Q) > Q$ as long as $G > Q$ (the normalizer condition). Hence in a nilpotent group, $N_G(P) = Q$ must be all of $G$, and all Sylow subgroups are normal. If $P,R$ are Sylow subgroups for different primes and are both normal, then $[P,R] \leq P \cap R = 1$ so $P$ and $R$ commute. Hence in a nilpotent group, $[x,y]=1$ if $\gcd(|x|,|y|)=1$.

Conversely, if $[x,y]=1$ whenever $\gcd(|x|,|y|)=1$, then each Sylow $p$-subgroup is normal (it is normalized by itself, and centralized by a supplement). Since $P$ and $R$ commute and are normal, $PR=RP = P \times R$, and doing this for all of the Sylows writes $G$ as a direct product of its (finitely many non-identity) Sylow subgroups. A direct product of nilpotent groups is nilpotent, and $p$-groups are nilpotent, so we get that $G$ is also nilpotent.