# Finding some explicit formula for $(ab)^n$ in any $a,b$ in a finite $p$-group.

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 $$H \leq G$$

• do you mean something like $(ab)^n=a^nb^n$ ? – Chinnapparaj R Dec 18 '18 at 9:29
• I know this is not true for all cases, I mean some thing $a^nb^nf(a,b)$ – A.Messab Dec 18 '18 at 10:09
• @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. – user1729 Dec 18 '18 at 11:48
• @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. – Tobias Kildetoft Dec 18 '18 at 11:57
• @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 $c_i$. – Derek Holt Dec 21 '18 at 19:41

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 :-)

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