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Socratica has a fantastic video on normal subgroups and quotient groups, but there’s one part of which I can’t convince myself.

Let $G$ be a commutative group under juxtaposition, let $N$ be a normal subgroup of $G$, and let the quotient group $G/N$ be equipped with the operation $\cdot$ for clarity. The members of $G/N$ are of the form $gN$, where $g\in G$, and the quotient group operation is defined by $$xN\cdot yN = zN \iff \forall g_1\in xN\ \ \forall g_2\in yN\ :\ g_1g_2\in zN$$

We know that the coset containing $g_1g_2=g_1g_2e$ is $g_1g_2N$ since $e\in N$. Therefore $\{g_1g_2N : g_1\in xN\wedge g_2\in yN\}\subseteq xN\cdot yN$. However, the video claims $xyN = xN \cdot yN$.

How does one go from $\{g_1g_2N\}\subseteq xN\cdot yN$ to $xyN = xN \cdot yN$? I tried to compare $G/N$ to $\Bbb Z/5\Bbb Z$, but that didn’t make this step any clearer.

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  • $\begingroup$ Because the commutativity of $G$ and the fact that $NN=N$. $\endgroup$ – A.G May 29 at 20:14
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Let $G$ be a group and $N$ a normal subgroup of $G$. $N$ being normal by definition means $gng^{-1} \in N$ for all $g \in G$ and for all $n \in N$.

This is equivalent to the statement $gNg^{-1} \subset N$ for all $g \in G$. And this is further equivalent to the statement $gNg^{-1} = N$ i.e. $gN = Ng$ for all $g \in G$.

Using this together with the fact that $N = NN$ (since $N$ is a subgroup), we have

$$(xN)(yN) = x(Ny)N = x(yN)N = xyNN = xyN.$$

A more direct way to see this is the following: Let $xN, yN \in G/N$ where $x, y \in G$. Then

$$ \begin{align} xN \cdot yN &= \{ xnyn : n \in N \} \\ & = \{ xy(y^{-1}ny)n: n \in N\} \\ & = \{ xy(y^{-1}n(y^{-1})^{-1})n: n \in N\} \\ &= \{ xyn'n: n,n' \in N\} \\ &= \{xy n'': n'' \in N \} \\ &= xyN, \end{align}$$

where in the second equality we used the fact that $e = yy^{-1}$ and in the fourth equality is where we used the normal property of $N$.

Addendum: Note that here, we don't even need $G$ to be commutative! Since it was given the fact that $G$ is a commutative group, we immediately have

$$xN \cdot yN = \{xnyn: n \in N \} = \{xynn: n \in N \} = xyN.$$

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