# How do I finish deriving this property of normal subgroups?

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.

• Because the commutativity of $G$ and the fact that $NN=N$. – A.G May 29 at 20:14

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

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