I am trying to prove the claim in the title.
I was able to do most of the work, but I still need some help.
I will show what I have written so far, and will highlight the parts in the proof that I wasn't sure about.
First direction:
Let $G$ be a group where $H\triangleleft G$ is a normal subgroup in $G$.
I'll define a relation: $\rho \subseteq G\times G$ as follows:
$g_1\sim g_2$ under $\rho$ iff $g_1g_2^{-1}\in H$.
I'll first show it is an equivalence relation:
1. Reflexivity: It is easy to see that $g\sim g$, since $gg^{-1}=e\in H$.
2. Symmetry: Suppose $g_1\sim g_2$, so by definition we have $g_1g_2^{-1}\in H$, so $g_1g_2^{-1}=h$ for some $h\in H$, and then $g_1g_2^{-1}=h\Rightarrow g_2g_1^{-1}=h^{-1}\Rightarrow g_2\sim g_1$.
3. Transitivity: Suppose $g_1\sim g_2$ and $g_2\sim g_3$, then:
$g_1g_2^{-1}=h$ and $g_2g_3^{-1}=k$, where $h,k\in H$.
Now from the last identity we have $g_2^{-1}=g_3^{-1}k^{-1}$, and then $g_1g_3^{-1}k^{-1}=h\Rightarrow g_1g_3^{-1}=hk\Rightarrow g_1\sim g_3$.
Now, I need to prove that this relation preserves the group structure, in the sense that the group operation is well defined on $\rho$'s equivalence classes, meaning, I need to prove that $g_1\sim g_2$ and $g_3\sim g_4$ implies that $g_1g_3\sim g_2g_4$, but couldn't make it work. I would love to get some help here…
Second direction:
Let $\rho$ be a congruence relation on a group $G$.
I'll define the set: $H=\left\{ g\in G| g\sim e \right\}$.
I'll first prove that this is a subgroup:
1. Of course, $e\sim e$, thus $e\in H$.
2. Suppose $h_1,h_2\in H$, then $h_1\sim e$ and $h_2\sim e$. since $\rho$ is a congruence relation, we have $h_1h_2\sim ee\Rightarrow h_1h_2\sim e \Rightarrow h_1h_2\in H$.
3. Suppose $h\in H$, then $h\sim e$. since $\rho$ is an equivalence relation, we have $h^{-1}\sim h^{-1}$, and since it is also a congruence relation, we get: $hh^{-1}\sim eh^{-1}\Rightarrow e\sim h^{-1}\Rightarrow h^{-1}\in H$.
Now, to show that $H$ is normal, I need to prove that $\forall g\in G$ and $\forall h\in H$, we have $ghg^{-1}\in H$.
This is also a part where I don't know how to proceed, and would really appreciate some guidance.
Best Answer
Second part:
Since $\sim$ is an equivalence relation, we have $g\sim g$ and $g^{-1}\sim g^{-1}$. Further, as $\sim$ is congruence, and $h\in H \implies h\sim e \implies gh\sim ge=g$ and so $ghg^{-1} \sim gg^{-1}=e$ for any $g\in G, h\in H$