## Equivalent characterization of group congruences

Let $G$ be a group. Assume that $\mathcal{P} = \{ A_i \ |\ i \in I \}$ is a partition of $G$ such that $\mathcal{P}$ is a group under the “quotient” operation as follows: to compute the product of $A_i$ and $A_j$ take any representatives $a_i \in A_i$ and $a_j \in A_j$ and let $A_iA_j$ be the element of $\mathcal{P}$ containing $a_ia_j$. (Assume this operation is well defined.)

Prove that the element $N$ of $\mathcal{P}$ containing 1 is a normal subgroup of $G$ and that the elements of $\mathcal{P}$ are the cosets of $N$, so that $\mathcal{P} = G/N$.

We will denote the $\mathcal{P}$ equivalence class of $x$ by $[x]$; hence $N = [1]$.

Lemma 1: Let $G$ and $\mathcal{P}$ be as described above. Then for all $g,x \in G$, $g[x]g^{-1} = [gxg^{-1}]$. Proof: $(\subseteq)$ we have $g[x]g^{-1} \subseteq [g][x][g^{-1}] = [gxg^{-1}]$. $(\supseteq)$ We have $g^{-1}[gxg^{-1}]g \subseteq [g^{-1}gxg^{-1}g] = [x]$, so that $[gxg^{-1}] \subseteq g[x]g^{-1}$. $\square$

Lemma 2: Let $G$ and $\mathcal{P}$ be as described above. Then for all $g,x \in G$, $g[x] = [gx]$. Proof: $(\subseteq)$ We have $g[x] \subseteq [g][x] = [gx]$. $(\supseteq)$ We have $g^{-1}[gx] \subseteq [g^{-1}gx] = [x]$, so that $[gx] \subseteq g[x]$. $\square$

First we show that $N \leq G$. Note that $N$ is nonempty since $1 \in N$. Now suppose $x,y \in N$. We have $1 = [x^{-1}x] = [x^{-1}][x]$, and so by the uniqueness of group inverses, $[x]^{-1} = [x^{-1}]$. Thus $[xy^{-1}] = [x][y]^{-1} = [1][1]^{-1} = [1] = N$, and so $xy^{-1} \in N$. By the Subgroup Criterion, $N$ is a subgroup of $G$.

Now we show that $N$ is normal in $G$. Let $g \in G$; now $gNg^{-1} = g[1]g^{-1} = [gg^{-1}] = [1] = N$ by Lemma 1.

Now we show that every $A \in \mathcal{P}$ is a coset of $N$. Let $A \in \mathcal{P}$ and $a \in A$. Now $A = [a] = a[1] = aN$, using Lemma 2. So $A$ is a coset of $N$.

Now we show that every coset of $N$ is in $\mathcal{P}$. Let $g \in G$. Since $\mathcal{P}$ is a partition, $g \in A$ for some $A \in \mathcal{P}$. Thus $A = [g] = g[1] = gN$, using Lemma 2, and so $gN \in \mathcal{P}$.

I like your notations. A typo: $xY^{-1}$ should be $xy^{-1}$.