# Commutator Identities in p-groups

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 !

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They both appear to be false in SmallGroup(64,32). – Derek Holt Jun 2 '12 at 14:38
Can you please give the formal definition of these groups? Or show me how to build a proper counter-example? Thanks! – joshua Jun 2 '12 at 17:49

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.

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Thanks!!!!!!!!!!! – joshua Jun 2 '12 at 18:19