Do these $p$-groups have the same nilpotency class?

Let $$G$$ be a $$p$$-group, $$\{e\}\not= H\subseteq G$$ be a subgroup of $$G$$ such that $$G' = H'$$. Is it true that $$c(G) = c(H)$$, where $$c(\cdot)$$ denotes the nilpotency class of a group?

• I suppose you assume $H$ to be non-trivial, otherwise $G = C_p$ for $p$ prime works as a counterexample. Sep 20, 2023 at 8:07
• Clearly $c(H)=1$ implies $c(G)=1$. Actually, it is also true that $c(H)=2$ implies $c(G)=2$. This is a consequence of the Hall-Witt identity $[[a,b^{-1}],c]^b \cdot [[b,c^{-1}],a]^c \cdot [[c,a^{-1}],b]^a = 1$, rewritten in the form $[[h_1,h_2],g]^{h_2^{-1}} \cdot [[h_2^{-1},g^{-1}],h_1]^c \cdot [[g,h_1^{-1}],h_2^{-1}]^{h_1} = 1$. If $H$ is 2-step nilpotent and $H'=G'$, both $[h_2^{-1},g^{-1}]$ and $[g,h_1^{-1}]$ are in $H'$ and hence both right-hand double brackets are trivial, so eventually $G$ centralizes $H'=G'$.
– YCor
Sep 20, 2023 at 8:33
• Finally this argument can be adapted to the general case, so I wrote an answer.
– YCor
Sep 20, 2023 at 9:01

This is true in an arbitrary nilpotent group.

Namely, for a nilpotent group $$G$$ and subgroup $$H$$ with $$G'=H'$$, let us check that $$G^{(i)}=H^{(i)}$$ for all $$i\ge 2$$ (lower central series. In particular, they have the same class (with the exception of $$c(G)=1$$, $$c(H)=0$$, i.e. $$G$$ nontrivial abelian $$H$$ trivial).

The case $$i=2$$ is the assumption $$G'=H'$$.

We use in general the Hall-Witt identity. $$[[a,b^{-1}],c]^b \cdot [[b,c^{-1}],a]^c \cdot [[c,a^{-1}],b]^a = 1$$ (with conventions $$x^y=y^{-1}xy$$ and $$[x,y]=x^{-1}y^{-1}xy$$).

Suppose that $$G^{(j)}=H^{(j)}$$ for all $$2\le j\le i$$. For $$g\in G$$, $$h\in H$$ and $$s\in H^{(i-1)}$$, this formula shows that $$[[h,s],g]$$ belongs to $$[[G,H],H^{(i-1)}][[H^{(i-1)},G],H].$$ We have $$[[H^{(i-1)},G],H]=[[G^{(i-1)},G],H]=[G^{(i)},H]=[H^{(i)},H]=H^{(i+1)}.$$ Also $$[[G,H],H^{(i-1)}]=[[H,H],H^{(i-1)}]$$ is contained, again using the Hall-Witt identity, in $$[[H^{(i-1)},H],H]=H^{(i+1)}$$. Hence $$[[h,s],g]\in H^{(i+1)}$$. Since such elements $$[h,s]$$ generate $$H^{(i)}=G^{(i)}$$, we deduce that $$G^{(i+1)}=[G,G^{(i)}]\subseteq [H,H^{(i)}]=H^{(i+1)}.$$

(By the way, this argument also works in the context of Lie rings.)

• Thank you! mathoverflow.net/questions/454950/… is another question. Sep 20, 2023 at 10:03
• You could also phrase your proof in terms of the three subgroups lemma (which follows from the Hall-Witt identity): Let $N$ be a normal subgroup of $G$ and $A,B,C \leq G$. If two among the subgroups $[A,B,C]$, $[B,C,A]$, $[C,A,B]$ are contained in $N$, then all of them are. ${}{}$ By the way, the problem appears in exercise 2.10 in Chap 4 of Suzuki, Group Theory II. Also in Corollary 1 of "Some sufficient conditions for a group to be nilpotent, P. Hall, Illinois J. Math. 2(4B): 787-801 (December 1958). DOI: 10.1215/ijm/1255448649". Sep 20, 2023 at 10:03