3
$\begingroup$

(Disprove) Suppose $H<G$. Suppose for every $g\in G$ and for every $h\in H$, $gh=h'g$ for some $h'\in H$. Then, for every $g\in G$ and for every $h\in H$, $gh=hg$.

You don't have to read below, but it is about why I am wondering this question. Suppose $H<G$ and consider following statements.

  1. $G$ is commutative.
  2. For every $g\in G$ and $h\in H$, $gh=hg$.
  3. For every $g\in G$ and $h\in H$, $gh=h'g$ for some $h'\in H$.
  4. $H\triangleleft G$ ($H$ is normal)

(My conjecture) $(1)\stackrel{\not\Leftarrow} \Rightarrow(2)\stackrel{\not\Leftarrow} \Rightarrow(3)\Leftrightarrow(4)$.

(1) $\Rightarrow$ (2) : trivial

(2) $\not\Rightarrow$ (1) : $G=$The symmetric group of square $=\{e,r,r^2,r^3,t_x,t_y,t_{AC},t_{BD}\}$, $H=\{e,r^2\}$.

(2) $\Rightarrow$ (3) : trivial

(3) $\not\Rightarrow$ (2) : That's what I am wondering

(3) $\Leftrightarrow$ (4) : \begin{align*} H\triangleleft G &\iff\forall g\in G,\: gHg^{-1}\subset H\\ &\iff\forall g\in G,\: \forall h\in H,\: ghg^{-1}\in H\\ &\iff\forall g\in G,\: \forall h\in H,\:\exists h'\in H\text{ s.t. }ghg^{-1}=h' \end{align*}

I want anyone who read to this point to verify if there is a counterexample for the above statment(Disprove) and if my conjecture is right or not. Thank you in advance!

$\endgroup$
3
$\begingroup$

This is false. Take $G=S_3, H=\langle (1\ 2\ 3)\rangle$. Since $H\trianglelefteq G$ we have $gH=Hg$ for all $g\in G$. However, $(1\ 2)(1\ 2\ 3)\ne (1\ 2\ 3)(1\ 2)$.

And yes, your conjecture is right.

$\endgroup$
1
  • $\begingroup$ As I comprehend, $|G|=6$ and $|H|=3$ in your example, so it is trivial that $H\triangleleft G$. Thus (4) holds. So we don't have to consider (more) complicated statement (3) and try to look at Cayley table of $S_3$. But (2) fails by $(12)\in G$ and $(123)\in H$. Thanks for your answer. $\endgroup$
    – shyzealot
    Dec 30 '19 at 3:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.