Group $G =\langle a, b\rangle$ with $a^{p^m}=b^{p^r}=1$ and $b^{-1}ab=a^n$ has commuting subgroups.

Asked Feb 09 '21 at 19:30

Active Feb 09 '21 at 20:57

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Here $n^{p^r}\equiv 1\pmod{p^m}$ and $p$ is prime.

We say $A, B\leq G$ commute if $AB =BA$.

I've shown that commuting subgroups is equivalent to $xy = y^tx^s$ for any $x, y\in G$ and some $t, s\in \mathbb{Z}$.

This is because one can restrict to cyclic groups when showing that any two subgroups commute.

Next I've noted that $G$ is a quotient of group of pairs $(t, s)\in \mathbb{Z}_{p^r}\times \mathbb{Z}_{p^m}$ with product $(t, s)*(t', s') := (t+t', n^{t'}s+s')$.

This is where number theory pops in. My question can be then phrased in terms of a system of equations:

\begin{cases} xt+yt'\equiv t+t'\pmod{p^r}\\ n^{yt'}\sum\limits_{i=0}^{x-1}n^{ti}s+\sum\limits_{i=0}^{y-1}n^{t'i}s' \equiv s+n^ts'\pmod{p^m}\end{cases} in positive integers $x, y$. This is where I'm stuck.

edited Feb 09 '21 at 20:57

asked Feb 09 '21 at 19:30

Jakobian

15,280

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What exactly is your question? – Derek Holt Feb 09 '21 at 20:00
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how to show the subgroups commute – Jakobian Feb 09 '21 at 20:05
Which subgroups are you trying to show commute? – Aryaman Maithani Feb 09 '21 at 20:25
@AryamanMaithani all of them – Jakobian Feb 09 '21 at 20:35
So you just want to show that $G$ is abelian, i.e., a commutative group? – Aryaman Maithani Feb 09 '21 at 20:43
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@AryamanMaithani It is possible for any pair of subgroups to commute in a nonabelian group (such as $Q_8$). – Derek Holt Feb 09 '21 at 22:12
It might help to note that $G$ looks like a semidirect product, especially when viewed as the group given by the group presentation $$\left\langle a,b\mid a^{p^m}, b^{p^r}, b^{-1}ab=a^n\right\rangle.$$ – Shaun Feb 09 '21 at 22:15
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These are usually called "permuting" subgroups, not "commuting". A subgroup $H$ of $G$ is permutable if for all subgroups $K$ of $G$, $HK=KH$; and we say $H$ and $K$ "permute" or are "permuting subgroups" if $HK=KH$. See for example, here – Arturo Magidin Feb 09 '21 at 22:24
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("Permutable subgroups" are also sometimes called "quasinormal", so you may also find the result you want using that term) – Arturo Magidin Feb 09 '21 at 22:33
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For odd $p$, the result is Satz 15 in Huppert's Uber das Produkt von paarweise vertauschbaren zyklischen Gruppen, Math Z. 58 (1953), 243-264, among other places. – Arturo Magidin Feb 09 '21 at 22:36
@DerekHolt, agreed. At the time the question was put up, the definition of commuting subgroups wasn't there and I thought that it meant that the elements of the two subgroups commute. – Aryaman Maithani Feb 10 '21 at 03:30
The result is false in dihedral groups, which has subgroups of order $2$ that do not permute, so it is true only for odd $p$. – Derek Holt Feb 10 '21 at 07:39

Group $G =\langle a, b\rangle$ with $a^{p^m}=b^{p^r}=1$ and $b^{-1}ab=a^n$ has commuting subgroups.

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