Let be a group and $N$ a normal subgroup of $G$ . Then we may define the quotient group $G / N$ to be the set of left cosets $g N$ of $N$ in $G$ , with the group operation that $g N + h N = (g h) N$ . This is well-defined if and only if $N$ is normal.

Proof

$N$ normal implies $G / N$ well-defined

Recall that $G / N$ is well-defined if "it doesn't matter which way we represent a coset": whichever coset representatives we use, we get the same answer.

Suppose $N$ is a normal subgroup of $G$ . We need to show that given two representatives $g_{1} N = g_{2} N$ of a coset, and given representatives $h_{1} N = h_{2} N$ of another coset, that $(g_{1} h_{1}) N = (g_{2} h_{2}) N$ .

So given an element of $g_{1} h_{1} N$ , we need to show it is in $g_{2} h_{2} N$ , and vice versa.

Let $g_{1} h_{1} n \in g_{1} h_{1} N$ ; we need to show that $h_{2}^{- 1} g_{2}^{- 1} g_{1} h_{1} n \in N$ , or equivalently that $h_{2}^{- 1} g_{2}^{- 1} g_{1} h_{1} \in N$ .

But $g_{2}^{- 1} g_{1} \in N$ because $g_{1} N = g_{2} N$ ; let $g_{2}^{- 1} g_{1} = m$ . Similarly $h_{2}^{- 1} h_{1} \in N$ because $h_{1} N = h_{2} N$ ; let $h_{2}^{- 1} h_{1} = p$ .

Then we need to show that $h_{2}^{- 1} m h_{1} \in N$ , or equivalently that $p h_{1}^{- 1} m h_{1} \in N$ .

Since $N$ is closed under conjugation and $m \in N$ , we must have that $h_{1}^{- 1} m h_{1} \in N$ ; and since $p \in N$ and $N$ is closed under multiplication, we must have $p h_{1}^{- 1} m h_{1} \in N$ as required.

$G / N$ well-defined implies $N$ normal

Fix $h \in G$ , and consider $h n h^{- 1} N + h N$ . Since the quotient is well-defined, this is $(h n h^{- 1} h) N$ , which is $h n N$ or $h N$ (since $n N = N$ , because $N$ is a subgroup of $G$ and hence is closed under the group operation). But that means $h n h^{- 1} N$ is the identity element of the quotient group, since when we added it to $h N$ we obtained $h N$ itself.

That is, $h n h^{- 1} N = N$ . Therefore $h n h^{- 1} \in N$ .

Since this reasoning works for any $h \in G$ , it follows that $N$ is closed under conjugation by elements of $G$ , and hence is normal.