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












3












$begingroup$



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.






abstract-algebra group-theory finite-groups p-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    do you mean something like $(ab)^n=a^nb^n$ ?
    $endgroup$
    – Chinnapparaj R
    Dec 18 '18 at 9:29






  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    @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 $(xy)^m=x^my^m[y, x]^{mchoose2}$ 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






  • 2




    $begingroup$
    @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






  • 2




    $begingroup$
    @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
















3












$begingroup$



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.






abstract-algebra group-theory finite-groups p-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    do you mean something like $(ab)^n=a^nb^n$ ?
    $endgroup$
    – Chinnapparaj R
    Dec 18 '18 at 9:29






  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    @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 $(xy)^m=x^my^m[y, x]^{mchoose2}$ 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






  • 2




    $begingroup$
    @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






  • 2




    $begingroup$
    @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














3












3








3


1



$begingroup$



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.






abstract-algebra group-theory finite-groups p-groups







share|cite|improve this question











$endgroup$





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.






abstract-algebra group-theory finite-groups p-groups




abstract-algebra group-theory finite-groups p-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:56









Shaun

9,789113684




9,789113684










asked Dec 18 '18 at 9:27









A.MessabA.Messab

527




527








  • 1




    $begingroup$
    do you mean something like $(ab)^n=a^nb^n$ ?
    $endgroup$
    – Chinnapparaj R
    Dec 18 '18 at 9:29






  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    @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 $(xy)^m=x^my^m[y, x]^{mchoose2}$ 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






  • 2




    $begingroup$
    @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






  • 2




    $begingroup$
    @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














  • 1




    $begingroup$
    do you mean something like $(ab)^n=a^nb^n$ ?
    $endgroup$
    – Chinnapparaj R
    Dec 18 '18 at 9:29






  • 1




    $begingroup$
    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






  • 3




    $begingroup$
    @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 $(xy)^m=x^my^m[y, x]^{mchoose2}$ 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






  • 2




    $begingroup$
    @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






  • 2




    $begingroup$
    @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








1




1




$begingroup$
do you mean something like $(ab)^n=a^nb^n$ ?
$endgroup$
– Chinnapparaj R
Dec 18 '18 at 9:29




$begingroup$
do you mean something like $(ab)^n=a^nb^n$ ?
$endgroup$
– Chinnapparaj R
Dec 18 '18 at 9:29




1




1




$begingroup$
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




$begingroup$
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




3




3




$begingroup$
@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 $(xy)^m=x^my^m[y, x]^{mchoose2}$ 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




$begingroup$
@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 $(xy)^m=x^my^m[y, x]^{mchoose2}$ 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




2




2




$begingroup$
@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




$begingroup$
@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




2




2




$begingroup$
@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




$begingroup$
@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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $$(xy)^m=x^my^m[y,x]^{mchoose2}$$ 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: $$x^my^m=(xy)^mc_2^{mchoose {2}}c_3^{mchoose {3}}cdots c_{m-1}^{mchoose {m-1}}c_m.$$ 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 :-)






share|cite|improve this answer









$endgroup$













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











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1 Answer
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oldest

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2












$begingroup$

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 $$(xy)^m=x^my^m[y,x]^{mchoose2}$$ 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: $$x^my^m=(xy)^mc_2^{mchoose {2}}c_3^{mchoose {3}}cdots c_{m-1}^{mchoose {m-1}}c_m.$$ 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 :-)






share|cite|improve this answer









$endgroup$













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
















2












$begingroup$

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 $$(xy)^m=x^my^m[y,x]^{mchoose2}$$ 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: $$x^my^m=(xy)^mc_2^{mchoose {2}}c_3^{mchoose {3}}cdots c_{m-1}^{mchoose {m-1}}c_m.$$ 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 :-)






share|cite|improve this answer









$endgroup$













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














2












2








2





$begingroup$

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 $$(xy)^m=x^my^m[y,x]^{mchoose2}$$ 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: $$x^my^m=(xy)^mc_2^{mchoose {2}}c_3^{mchoose {3}}cdots c_{m-1}^{mchoose {m-1}}c_m.$$ 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 :-)






share|cite|improve this answer









$endgroup$



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 $$(xy)^m=x^my^m[y,x]^{mchoose2}$$ 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: $$x^my^m=(xy)^mc_2^{mchoose {2}}c_3^{mchoose {3}}cdots c_{m-1}^{mchoose {m-1}}c_m.$$ 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 :-)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 10:54









user1729user1729

17.5k64193




17.5k64193












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


















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
















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




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


















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