Quotient by subgroup is well defined if and only if subgroup is 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.


$~$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.