We mentioned previously that given a subgroup \(H\) of \(G\text{,}\) we'd like to use \(H\) to get at some understanding of \(G\)'s entire structure. Recall that we've defined \(G/H\) to be the set of all left cosets of \(H\) in \(G\text{.}\) What we'd like to do now is equip \(G/H\) with some operation under which \(G/H\) is a group! The natural way to do this would be to define multiplication on \(G/H\) by

\begin{equation*}
(aH)(bH)=abH \text{ for all } a,b\in G.
\end{equation*}
Ok, so let's do that! But wait: we're defining this operation by referring to coset representatives, so we'd better make sure our operation is well-defined. Only it sadly turns out that in general this operation is *not* well-defined. For example:

#####
Example8.1.1

Let \(H=\langle (12)\rangle\) in \(S_3\text{,}\) and let \(a=(13)\) and \(b=(132)\text{.}\) Let \(x=aH\) and \(y=bH\) in \(S_3/H\text{.}\) Then using the above-defined operation on \(G/H\) we'd have

\begin{equation*}
xy=(aH)(bH)=abH=(13)(132)H=(23)H.
\end{equation*}
But also \(x=(123)H\) and \(y=(23)H\text{,}\) so we'd also have

\begin{equation*}
xy=((123)H)((23)H))=(123)(23)H=(12)H=H\neq (23)H.
\end{equation*}
So this operation isn't well-defined.

The question for us now becomes: what conditions must hold for \(H\) in \(G\) in order for operation

\begin{equation*}
(aH)(bH)=abH
\end{equation*}
on \(G/H\) to be well-defined? It turns out that this operation is well-defined exactly when \(H\) is normal in \(G\text{!}\) We state this in the following theorem:

#####
Theorem8.1.2

Let \(H\leq G\text{.}\) Then the operation

\begin{equation*}
(aH)(bH)=abH
\end{equation*}
on \(G/H\) is well-defined if and only if \(H \unlhd G\text{.}\)

##### Proof

First, assume that the described operation is well-defined. Let \(a\in G\text{;}\) we want to show that \(aH=Ha\text{.}\)

Well, let \(x\in aH\text{.}\) Then

\begin{equation*}
(xH)(a^{-1}H)=xa^{-1}H
\end{equation*}
and, since \(xH=aH\text{,}\)

\begin{equation*}
(xH)(a^{-1}H)=(aH)(a^{-1}H)=aa^{-1}H=H.
\end{equation*}
Since our operation on \(G/H\) is well-defined, this means that \(xa^{-1}H=H\text{,}\) so \(xa^{-1}\in H\text{;}\) thus, \(x\in Ha\text{.}\) We conclude that \(aH\subseteq Ha\text{.}\) The proof that \(Ha\subseteq aH\) is similar. So \(aH=Ha\text{,}\) and thus, since \(a\in G\) was arbitrary, \(H\) is normal in \(G\text{.}\)

Conversely, assume \(H\unlhd G\text{.}\) Let \(a_1,a_2,b_1,b_2\in G\) with \(a_1H=a_2H\) and \(b_1H=b_2H\text{.}\) We want to show that then \(a_1b_1H=a_2b_2H\text{,}\) that is, that \((a_2b_2)^{-1}a_1b_1\in H\text{.}\) Well, since \(a_1H=a_2H\) we have \(a_2^{-1}a_1\in H\text{.}\) So

\begin{equation*}
(a_2b_2)^{-1}a_1b_1=b_2^{-1}(a_2^{-1}a_1)b_1 \in b_2^{-1}Hb_1.
\end{equation*}
Next, since \(H\unlhd G\text{,}\) we have \(Hb_1=b_1H\text{,}\) so \(b_2^{-1}Hb_1=b_2^{-1}b_1H\text{;}\) but since \(b_1H=b_2H\text{,}\) we have \(b_2^{-1}b_1\in H\text{,}\) so \(b_2^{-1}b_1H=H\text{.}\) Thus, \((a_2b_2)^{-1}a_1b_1\in H\text{,}\) as desired.

#####
Definition8.1.3

When \(H\unlhd G\text{,}\) the well-defined operation \((aH)(bH)=abH\) on \(G/H\) is called *left coset multiplication*.

It is clear that normal subgroups will be very important for us in studying group structures. We therefore spend some time looking at normal subgroups before returning to equipping \(G/H\) with a group structure.