(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.
- $G$ is commutative.
- For every $g\in G$ and $h\in H$, $gh=hg$.
- For every $g\in G$ and $h\in H$, $gh=h'g$ for some $h'\in H$.
- $H\triangleleft G$ ($H$ 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.