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.