Is it trivial that given a finite p-group $G$ , then: $ [G^p ,G^p] \subseteq [G,G]^p [ [G,G ] , [G,G] ] $ ? What about $ [G^p ,[G,G]] \cap [G,G] \subseteq [G,G]^p [ [G,G ] , [G,G] ] $ ?
Is it true at all?
Help is needed !
Thanks in advance
Is it trivial that given a finite p-group $G$ , then: $ [G^p ,G^p] \subseteq [G,G]^p [ [G,G ] , [G,G] ] $ ? What about $ [G^p ,[G,G]] \cap [G,G] \subseteq [G,G]^p [ [G,G ] , [G,G] ] $ ?
Is it true at all?
Help is needed !
Thanks in advance
SmallGroup(64,32) has the Power-Conjugate presentation:
G.1^2 = G.4, G.2^G.1 = G.2 * G.3, G.3^G.1 = G.3 * G.5, G.4^G.2 = G.4 * G.5, G.4^G.3 = G.4 * G.6, G.5^G.1 = G.5 * G.6
The convention here is that a generator has order 2 if not stated otherwise, and two generators commute if not otherwise specified.
It is easy to see that the derived group $[G,G]$ is elementary abelian of order 8, and generated by G.3, G.5, G.6. So $[G,G]^2 = [[G,G],[G,G]]=1$.
Now $G^2$ contains G.1^2 = G.4 and also
(G.1*G.2)^2 = G.4*G.2^G.1*G.2 = G.4*G.2*G.3*G.2 = G.4*G.3,
which does not commute with G.4, so $[G^2,G^2] \ne 1$.
Also $[G^2,[G,G]]$ contains [G.4,G.3], which is nontrivial.