(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!


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.

  • $\begingroup$ 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

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