8.1: Motivation
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:
Example \(\PageIndex{1}\)
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:
Theorem \(\PageIndex{1}\)
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 \(a∈G\); we want to show that \(aH=Ha\).
Well, let \(x∈aH\). Then
\((xH)(a^{−1}H)=xa^{−1}H\)
and, since \(xH=aH\)
\((xH)(a^{−1}H)=(aH)(a^{−1}H)=aa^{−1}H=H\).
Since our operation on \(G/H\) is well-defined, this means that \(xa^{−1}H=H\), so \(xa^{−1}∈H\); thus, \(x∈Ha\). We conclude that \(aH⊆Ha\). The proof that \(Ha⊆aH\) is similar. So \(aH=Ha\), and thus, since \(a∈G\) was arbitrary, \(H\) is normal in \(G\).
Conversely, assume \(H⊴G\). Let \(a_1,a_2,b_1,b_2∈G\) with \(a_1H=a_2H\) and \(b_1H=b_2H\). We want to show that then \(a_1b_1H=a_2b_2H\), that is, that \((a_2b_2)^{−1}a_1b_1∈H\). Well, since \(a_1H=a_2H\) we have \(a_2^{−1}a_1∈H\). So
\((a_2b_2)^{−1}a_1b_1=b_2^{−1}(a_2^{−1}a_1)b_1∈b_2^{−1}Hb_1\).
Next, since \(H⊴G\), we have \(Hb_1=b_1H\), so \(b_2^{−1}Hb_1=b^{−1}_2b_1H\); but since \(b_1H=b_2H\), we have \(b_2^{−1}b_1∈H\), so \(b^{−1}_2b_1H=H\). Thus, \((a_2b_2)^{−1}a_1b_1∈H\), as desired.
Definition: Left Coset Multiplication
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.