4

Let $G$ be a group with two generators $x,y$ such that $x^{p^m}=1$, $y^{p^r}=1$, $y^{-1}xy=x^m$ where $m^{p^r}\equiv 1 \pmod {p^m}$, $p$ a prime.
Show that if $H_1$ and $H_2$ are subgroups of $G$, then $H_1H_2=H_2H_1$.

I can not see too much.

The only thing I can see is

$$y^a x^b y^c x^d=y^{a+c}x^{d+b m^c}.$$

So for a fixed element $y^a x^b$, we know that

$$ (y^a x^b)^k=y^{ka} x^{b(1+m^a+\cdots+m^{ka})}. $$

Arturo Magidin
  • 417,286
stlinex
  • 1,011
  • Here $G$ looks like a semidirect product of two cyclic groups. – Shaun Dec 04 '24 at 17:21
  • 1
    Do you really mean the $m$ from the order of $x$ and the $m$ from the conjugation action to be the same? – Arturo Magidin Dec 04 '24 at 17:22
  • https://math.stackexchange.com/q/160870/104041 – Shaun Dec 04 '24 at 17:27
  • I'm not sure it is; I'm just saying it looks like one. – Shaun Dec 04 '24 at 17:33
  • 1
    @Shaun: It is definitely a semidirect product: you have $y$ acting on $\langle x\rangle$ via an automorphism of order dividing the order of $y$. That said, your link does not seem to have anything to do with the question. – Arturo Magidin Dec 04 '24 at 17:35
  • Does the link not support my semidirect product claim, @ArturoMagidin? I thought it was obvious . . . – Shaun Dec 04 '24 at 17:41
  • 1
    @Shaun: But why does writing it a semidirect product (which is already being done, essentially, with normal forms) address the permutability of subgroups? – Arturo Magidin Dec 04 '24 at 17:46
  • I don't know. I thought it might help, @ArturoMagidin, that's all. – Shaun Dec 04 '24 at 17:47
  • 2
    Basically, it is both necessary and sufficient to prove this for cyclic subgroups; that means that you need to show that there always exist $k$ and $\ell$ such that $(y^ax^b)(y^cx^d) = (y^cx^d)^k(y^ax^b)^{\ell}$. – Arturo Magidin Dec 04 '24 at 17:48
  • @ArturoMagidin I see this in Jacobson, Basic Algebra I, Exercise 8.2.2. And I double check it. – stlinex Dec 05 '24 at 03:06

0 Answers0