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If $G$ is a finite $p-$group, let $a$ and $b$ any two elements from $G$.

Is there any formula for $(ab)^n$ involving $a^nb^n$ for any natural number $n$? That is, some formula like $(ab)^n = a^nb^nf(a,b)$ for some function of $a$ and $b$.

Please not those "modulo" some subgroups formulas! Like $$ (ab)^n \equiv a^nb^n \pmod{H}, $$ where $H \leq G$

Thanks in advance.

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    $\begingroup$ do you mean something like $(ab)^n=a^nb^n$ ? $\endgroup$ – Chinnapparaj R Dec 18 '18 at 9:29
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    $\begingroup$ I know this is not true for all cases, I mean some thing $a^nb^nf(a,b)$ $\endgroup$ – A.Messab Dec 18 '18 at 10:09
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    $\begingroup$ @A.Messab It is difficult to work out what you want. In $p$-groups, and more generally nilpotent groups, the formulaes usually involve commutators. For example, if $G$ is nilpotent of class at most two then the identity $(xy)^m=x^my^m[y, x]^{m\choose2}$ holds. Which is really pretty. You can generalise this to groups of higher class, but the cost here is more commutators. $\endgroup$ – user1729 Dec 18 '18 at 11:48
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    $\begingroup$ @user1729 There is a general formula due to Hall and Petrescu (I don't have the book by Hall on hand to see if that is the one). It is referred to as the Hall-Petrescu formula in Berkovich's book on $p$-groups. $\endgroup$ – Tobias Kildetoft Dec 18 '18 at 11:57
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    $\begingroup$ @hardmath well I thought the answer was just no, but that does not mean that the question is unclear! The general Hall-Petrescu formula is probably unhelpful because it involves unknown elements $c_i$. $\endgroup$ – Derek Holt Dec 21 '18 at 19:41
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There are some formulas* and they involve (often unknown) commutators. Sometimes there are particularly nice formulas, but they hold in $p$-groups and for things of the form $(ab)^{p^n}$, so they do not hold for arbitrary powers.

Some example formuals are:

  1. If $G$ is nilpotent of class at most two then the identity $$(xy)^m=x^my^m[y,x]^{m\choose2}$$ holds. Which is really pretty. You can generalise this, via the Hall-Petresco formula, to groups of higher class, but the cost here is more commutators.

  2. If $G$ is any group then we have the Hall-Petresco formula, as mentioned in the comments: $$x^my^m=(xy)^mc_2^{m\choose {2}}c_3^{m\choose {3}}\cdots c_{m-1}^{m\choose {m-1}}c_m.$$ Here each $c_i$ is contained in the $i^{th}$ subgroup of the decending central series for the group $G$. See Section 12.3 of M. Hall (1959), The theory of groups, Macmillan, MR 0103215. This book also contains some useful formulas for regular $p$-groups. Note that it is Philip Hall after whome the formula is named, while Marshall Hall wrote a book (and did other stuff too!).

You can find other formulas in the first section of the book C. Leedham-Green, S. McKay (2002) The Structure of Groups of Prime Power Order, Oxford University Press, and also in the book Y. Berkovich (2008) Groups of Prime Power Order Volume 1 (Appendix 1 of this book proves the Hall-Petresco formula).

*or formulae, if you want to be correct but also sound slightly pretentious :-)

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  • $\begingroup$ Thanks this is so helpful $\endgroup$ – A.Messab Jan 12 at 1:06

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