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Section 1.12 Quotient groups
Given a subgroup \(H\) of \(G\) we would like to put a group structure on the coset space \(G/H\text{:}\) indeed, what could be more natural than to define our binary operation as
\begin{equation}
gH\cdot g'H=(gg')H\text{?}\tag{1.12.1}
\end{equation}
It would seem the relevant group operations would follow readily from this definition. The issue is that the proposed operation is not always well defined! That is, if \(gH=g'H\) and \(kH=k'H\) for elements \(g,g',k,k'\in G\text{,}\) it is not necessarily the case that \((gk)H=(g'h')H\text{.}\)
Indeed, continuing with
Example 1.11.6 , where
\(G=D_4\) and
\(H=\angvec{rs}\text{,}\) we have
\begin{align*}
rH \amp =(r^2s)H\amp \\
r^3H \amp =sH
\end{align*}
and yet
\begin{align*}
(rr^3)H \amp =H \amp \\
((r^2s)s)H\amp =r^2H\ne H\text{.}
\end{align*}
It turns out that magic property a subgroup
\(H\) needs to satisfy in order for the operation
(1.12.1) to be well defined is that
\(H\) be a
normal subgroup .
Definition 1.12.1 . Normal subgroup.
A subgroup \(N\) of the group \(G\) is called normal if
\begin{align*}
gNg^{-1} \amp = N
\end{align*}
for all \(g\in G\text{.}\) We write \(N\normalin G\) to indicate that \(N\) is a normal subgroup of \(G\text{.}\)
Theorem 1.12.2 . Normal subgroups.
Let \(N\) be a subgroup of the group \(G\text{.}\) The following statements are equivalent.
\(\displaystyle N_G(N)=G\)
\(gN=Ng\) for all
\(g\in G\text{.}\)
The operation
\begin{align*}
G/N \times G/N\amp \rightarrow G/N\\
(gN, hN) \amp \longmapsto (gh)N
\end{align*}
is well defined.
Proof. The equivalence
\((1)\iff (2)\) follows directly from the definition of normality and
\(N_G(N)\text{.}\)
The equivalence
\((1)\iff (3)\) is left as an exercise.
We prove \((3)\implies (4)\text{.}\) Assume \(gN=Ng\text{.}\) Assume \(gN=g'N\) and \(hN=h'N\text{.}\) We will show that
\begin{equation*}
(gh)N=(g'h')N\text{.}
\end{equation*}
Since \(gN=g'N\) and \(hN=h'N\text{,}\) we have \(g=g'n\) and \(h=h'n'\) for some elements \(n,n'\in N\text{,}\) and thus
\begin{equation*}
gh=g'nh'n'\text{.}
\end{equation*}
Next since \(hN=Nh\) (assuming (3)), we have \(nh'=h'n''\) for some \(n''\in N\text{,}\) yielding
\begin{equation*}
gh=g'h'n'' n'\in (g'h')N\text{.}
\end{equation*}
It follows from
Theorem 1.11.7 that
\((gh)N=(g'h')N\text{,}\) as desired.
The implication
\((4)\implies (3)\) is left as an exercise.
Theorem 1.12.3 . Quotient groups.
Let \(N\) be a normal subgroup of the group \(G\text{.}\)
The pair \((G/N, \cdot)\text{,}\) where \(\cdot\) is the operation
\begin{align*}
G/N\times G/N \amp \rightarrow G/N\\
(gN, g'N) \amp \rightarrow gN\cdot g'N=(gg')N \text{,}
\end{align*}
is a group.
The group identity of
\(G/N\) is the coset
\(N\text{.}\) Given coset
\(gN\in G/N\) its group inverse is the coset
\(g^{-1}N\text{.}\)
The map
\(\pi\colon G\rightarrow G/N\) defined as
\(\pi(g)=gN\) is a surjective group homormorphism satisfying
\(\ker\pi=N\text{.}\)
Proof.
Example 1.12.4 . Normality in \(D_4\) .
Decide whether the given subgroup of
\(D_4\) is normal. Verify your answer by checking multiple conditions of the equivalence in
Theorem 1.12.2 .
\(\displaystyle H=\angvec{rs}\)
\(\displaystyle H=\angvec{r}\)
\(\displaystyle H=\angvec{r^2}\)
Example 1.12.5 . Abelian groups.
Show that an abelian group
\(G\) every subgroup is normal.
Solution .
Let \(G\) be abelian. Given a subgroup \(H\leq G\text{,}\) for any \(h\in H\) and \(g\in G\) we have
\begin{align*}
ghg^{-1} \amp =gg^{-1}h \amp (G \text{ abelian})\\
\amp =h\in H\text{.}
\end{align*}
It follows that \(gHg^{-1}H\) for all \(g\in G\) and hence that \(H\) is normal.
Kernels of group homomorphisms are normal subgroups. In fact, as we show below, a subgroup is normal if and
only if it is the kernel of a group homomorphism.
Corollary 1.12.6 . Normality and kernels.
Let \(N\) be a subgroup of \(G\text{.}\) The following statements are equivalent.
\(N=\ker\phi\) for some group homomorphism
\(\phi\colon G\rightarrow H\text{.}\)
Example 1.12.7 . Normality in \(S_4\) .
Determine whether the given subgroup is normal in \(S_4\text{.}\) If it is, identify the corresponding quotient group with a familiar group.
\(\displaystyle H=\angvec{(1234)}\)
\(\displaystyle H=\angvec{(12)(34),(13)(24)}\)
