# Quotient by subgroup is well defined if and only if subgroup is normal

https://arbital.com/p/quotient_by_subgroup_is_well_defined_iff_normal

by Patrick Stevens Jun 17 2016 updated Jun 20 2016

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

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

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.