Relationship between the center and commutator subgroup of a group of nilpotency class 3 I am new to the theory of nilpotent groups. I am dealing with the nilpotent groups of class 3. I want to know is there any relationship between the commutator subgroup $[G, G]$ of the group $G$ and the center $Z(G)$ of the group $G$.  I have a strong belief that $[G, G]^{3}\subseteq Z(G)$ when 3 divides the order of the group $G$. But I could not find any literature for that. Can anyone please help me with this. Thank you, in advance.
 A: Your "strong belief" is false. Consider the $4 \times 4$ Heisenberg group
$$H_4(\mathbb{F}_p) = \left[ \begin{array}{cccc} 1 & \mathbb{F}_p & \mathbb{F}_p & \mathbb{F}_p \\ 0 & 1 & \mathbb{F}_p & \mathbb{F}_p \\ 0 & 0 & 1 & \mathbb{F}_p \\ 0 & 0 & 0 & 1 \end{array} \right]$$
over a prime $p \neq 3$, which is nilpotent of class $3$. Since the order of this group is not divisble by $3$, the map $g \mapsto g^3$ is a bijection on every cyclic subgroup and hence a bijection on every subgroup, so $[G, G]^3 \subseteq Z(G)$ iff $[G, G] \subseteq Z(G)$. The center consists of matrices which have zero entries except in the top right and has order $p$. But the commutator subgroup consists of matrices which have zero entries in the superdiagonal and has order $p^3$.
A: This does not hold even in a $3$-group.
Note that in any group $G$, using the commutator convention $[a,b]=a^{-1}b^{-1}ab$, we have
$$[x^r,y^s] \equiv [x,y]^{rs}[x,y,x]^{s\binom{r}{2}}[x,y,y]^{r\binom{s}{2}}\pmod{G_4}$$
where $G_n$ is the $n$th term of the lower central series of $G$, and $\binom{a}{2}=\frac{a(a-1)}{2}$ for any integer $a$. So you would be asserting that if $G$ is of class three, then for all $y\in G$ and $c\in [G,G]$,
$$[c^3,y] = [c,y]^3[c,y,c]^{\binom{3}{2}}=[c,y]^3$$
is trivial. In particular, $[x,y,x]$ and $[x,y,y]$ would both have to be of exponent $3$ for any $x,y\in G$.
This is definitely not true.
By a result of Remeslennikov and of Jónsson (independently), a variety of nilpotent groups of class $3$ is determined by a 4-tuple of nonnegative integers $(m,n,p,q)$  satisfying

*

*$n|\frac{m}{\gcd(m,2)}$;

*$p|m$;

*$q|\frac{m}{\gcd(m,6)}$;

*$p|3q$;

corresponding to the identities $$x^m=[x,y]^n=[x,y,z]^p=[x,y,y]^q=[x,y,z,w]=1.$$
(See here for citations). That means that no other identities will hold universally for groups of nilpotency class $3$. In particular, if we require $q\gt 3$ then it is not the case that $[c,y]^3=1$ for all $c\in [G,G]$. For example, we can take
$m=n=p=27$ and $q=9$. Then the corresponding relatively free group or rank 2 would not satisfy the identity you want.
This would be the group generated by $x$ and $y$, consisting of all elements of the form
$$x^a y^b [y,x]^c[y,x,x]^d[y,x,y]^e,\qquad a,b,c\in\mathbb{Z}/27\mathbb{Z},\qquad d,e\in\mathbb{Z}/9\mathbb{Z}$$
with multiplication given by
$$\Bigl(x^a y^b [y,x]^c[y,x,x]^d[y,x,y]^e\Bigr) \Bigl(x^{\alpha} y^{\beta} [y,x]^{\gamma}[y,x,x]^{\delta}[y,x,y]^{\epsilon}\Bigr)\\ =
x^{a+\alpha} y^{b+\beta} [y,x]^{c+\gamma+b\alpha} [y,x,x]^{d+\delta+c\alpha + b\binom{\alpha}{2}}[y,x,y]^{e+\epsilon+c\beta+b\alpha\beta+\alpha\binom{b}{2}}$$
with exponents taken modulo the appropriate power of $3$. In this group, neither $[y,x,y]$ nor $[y,x,x]$ are of exponent $3$, hence $[y,x]^3$ is not central.
Your "strong suspicion" only holds if $G_3$ has exponent $3$, and for $3$-groups of class $3$, the exponent of $G_3$ can be arbitrarily large.
