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Given a group G, and it's lower central series denoted by {$G_i$}. I want to prove that every set of elements ,$g_1,...,g_k \in G$, and for every natural $m$, there exists $h_1\in G_1$,...,$h_m\in G_m$ , such that:

$g_1^m$...$g_k^m$=$h_1^\binom{m}{1}$...$h_m^\binom{m}{m}$

I have heard that this is supposed to be challenging, but I'm really stumped for even a direction. Is proof by induction on $m$ the way to go?

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  • $\begingroup$ That looks like a generalization of the Hall-Petrescu formula, which relates $(xy)^m$ to $x^my^m$ via such powers of elements in the central series. $\endgroup$ – Tobias Kildetoft Mar 2 '17 at 9:59
  • $\begingroup$ Do you know of a refernce to somewhere elaborating more on the formula and it's derivation? $\endgroup$ – Keen-ameteur Mar 2 '17 at 10:37
  • $\begingroup$ Wikipedia seems to have several references (also, apparently, it should be Petresco rather than Petrescu) $\endgroup$ – Tobias Kildetoft Mar 2 '17 at 10:42
  • $\begingroup$ See the same question here. $\endgroup$ – Dietrich Burde Mar 4 '17 at 12:02

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