# Counterexamples for $gh=h'g$ for some $h'\in H$ doesn't implies $gh=hg$.

(Disprove) Suppose $$H. Suppose for every $$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 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, $$(1\ 2)(1\ 2\ 3)\ne (1\ 2\ 3)(1\ 2)$$.
• 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. Dec 30 '19 at 3:17