1.6. Normal subgroups
Normal subgroups are special subgroups which exhibit properties of interest for when we go on to later define the idea of quotient groups, a concept we have touched upon slightly in considering \(\mathbb{Z}/2\mathbb{Z}\) and other modulo groups. They are a bit abstract at first, since they have to do with cosets. Once you work with normal subgroups for a bit, it will s eventually click and the reasoning behind their definitions becomes clear.
Let \(G\) be a group and suppose \(H\) is a subgroup of \(G\). We say that \(H\) is normal if and only if for every \(g \in G\), we have that \(Hg = gH\). We denote such a relation as \(H \unlhd G\). \noindent We make two remarks here.
- Commutative Groups. Note that if \(G\) is commutative, then \(H\), a subgroup of \(G\), is also commutative. In fact, \(H\) commutes with all elements of \(G\). That is, if \(H = \{h_1, h_2, \dots \}\) then
for all \(g \in G\). Thus what we're trying to say here is if \(G\) is commutative, every subgroup \(H\) of \(G\) is normal. * Set Equality. If \(H\) is normal to \(G\), then \(gH = Hg\) all \(g \in G\). Be careful with this equation, since what this is not saying is that \(gh=hg\) for all \(g\in G\) and \(h \in H\); that would imply commutativity, and it may be the case that \(G\) and \(H\) are not commutative groups. That is, the above equation is set equality, not term-by-term equality.
What this does say, however, is if \(gH = Hg\), then for each \(g\in G\), and for every \(h_1 \in H\), there exists an \(h_2 \in H\) such that
Note here that commutative groups satisfy this because in their case, \(h_1 = h_2\) satisfies the equation.
Since our current definition of normality would be exhausting to use directly if we wanted to check if a subgroup is normal, we have the following theorem that helps us check for normality.
Let \(G\) be a group and \(H\) a subgroup of \(G\). The following are equivalent:
- 1. \(H \normal G\) for all \(g \in G\)
- 2. \(gHg^{-1} = H\) for all \(g \in G\)
- 3. \(gHg^{-1} \subset H\) for all \(g\in G\).
- 4. \((Hg)(Hh) = H(gh)\) for all \(g, h \in G\)
We'll prove this by producing a chain of imply statements that can traverse in both directions. Let \(G\) be a group and \(H\) be a subgroup.
\noindent \(\mathbf{(1 \iff 2)}\) If \(H \normal G\), then \(gH = Hg\) for all \(g \in G\). Multiplying on the left by \(g^{-1}\), we then see that \(gHg^{-1} = H\) for all \(g \in G\).
Proving the reverse direction, if \(gHg^{-1} = H\) for all \(g \in G\) then \(gH = Hg\) for all \(g \in G\), which means that \(H\) is normal by defintion.
\noindent \(\mathbf{(2 \iff 3)}\) If \(gHg^{-1} = H\) for all \(g \in G\) then it is certainly true that \(gHg^{-1} \subset H\) for all \(g \in G\).
Now we prove the other direction. Suppose \(gHg^{-1} \subset H\) for all \(g \in G\). Then
by multiplying on the right by \(g\) and on the left by \(g^{-1}\). However, since we have assumed (3) is true we know that
By the above equations we then have that \(H = g^{-1}Hg\), and multiplying by \(g^{-1}\) on the right and \(g\) on the left yields that \(H = gHg^{-1}\) as desired.
\noindent\(\mathbf{(2 \iff 4)}\) Suppose (2). Then observe that \(gHg^{-1} = H \implies gH = Hg\) for all \(g \in G\). Therefore for \(h \in G\),
In the first step we used associativity and in the second step we used the fact that \(gH = Hg\).
To prove the other direction, suppose \((Hg)(Hh) = H(gh)\) for all \(g, h \in G\). Let \(h = e\). Then To show a subgroup \(H\) of \(G\) is normal, condition (3) of this theorem generally the fastest and easy way to take advtange of. It is usually the least complicated one to show. \ \ \noindent Example. \ Consider the group \(GL_n(\mathbb{R})\) and its subgroup \(SL_n(\mathbb{R})\). It turns out that \(SL_n(\mathbb{R}) \normal GL_n(\mathbb{R})\), which we will show using condition (3).
Let \(A \in GL_n(\mathbb{R})\) and suppose \(T \in SL_n(\mathbb{R})\). We must show that \(ATA^{-1} \in SL_n(\mathbb{R})\) for all \(A \in GL_n(\mathbb{R})\) and \(T \in SL_n(\mathbb{R})\). Observe that
where we used the basic properties of the determinant for the
calculation. Since \(\det(ATA^{-1}) = 1\), we have that \(ATA^{-1}
\in SL_n(\mathbb{R})\) for all \(A\) and \(T\) in \(GL_n(\mathbb{R})\)
and \(SL_n(\mathbb{R})\), respectively. Therefore \(SL_n(\mathbb{R})\)
is normal to \(GL_n(\mathbb{R})\).
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Example.
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One important example is the following: for any group homomorphism
\(\phi\) between two groups \(G\) and \(G'\), recall that
\(\mbox{ker}(\phi)\) is a subgroup of \(G\). However, we also have
that \(\mbox{ker}(\phi) \normal G\), which we'll show as follows.
Let \(G, G'\) be groups and \(\phi: G \to G'\) be a group homomorphism. Then \(\ker(\phi) \normal G\).
We need to show that for all \(g \in G\), \(h \in \mbox{ker}(\phi)\) that \(ghg^{-1} \in \mbox{ker}(\phi)\). Thus observe that
Since \(\phi(ghg^{-1}) = 0\), we thus see that \(ghg^{-1} \in
\mbox{ker}(\phi)\) for all \(g \in G\) and \(h \in \mbox{ker}(\phi)\),
which proves \(\mbox{ker}(\phi) \normal G\).
Another important example of normality is the fact that the center of a group \(Z(G)\) is normal to \(G\) for any group \(G\).
Let \(G\) be a group. Then \(Z(G) \normal G\).
Recall that \(Z(G)\) is a subgroup of \(G\), consisting of all the elements of \(G\) which commute with every element in \(G\). More precisely,
Now for any \(g \in G\) and \(z \in Z(G)\), we have that \(gzg^{-1} = gg^{-1}z = z\), since \(z\) commutes with all elements of \(G\). Therefore \(gzg^{-1} \in Z(G) \implies gZ(G)g^{-1} \subset Z(G)\). By the previous theorem, we can conclude that \(Z(G) \normal G\) as desired.
Next, we introduce a small theorem that allows us to quickly and easily identify if a subgroup \(H\) of \(G\) is normal.
If \(G\) is a group and \(H\) is a subgroup, and \([G:H] = 2\), then \(H \normal G\).
Since \(G\) has two right (and equivalently two left) cosets, we see that they must be of the form \(H\) and \(Hg\) where \(g \in G\setminus H\) (that is, all of the elements of \(G\) which are not in \(H\)).
As we said before, there are equivalently two left cosets \(H\) and \(gH\) where \(g \in G\setminus H\). Since the cosets partition \(G\), we see that for any \(g \in G\setminus H\) two partitions of \(G\) are
Since these partition the same set we see that \(gH = Hg\) for all \(g \in G\setminus H\). Note that we already know that for \ \(g \in H\), \(Hg = H\) and \(gH = H\) so \(gH = Hg\). Therefore, we have all together that \(Hg = gH\) for all \(g \in G\).
\noindent In working with normal subgroups, one may form the following questions. \
\textcolor{ForestGreen}{Q: If \(K\) is a normal subgroup of \(H\) and \(H\) is a normal subgroup of \(G\), is \(K\) normal to \(G\)?} \
A: Not always. If \(H \normal K\), then \(khk^{-1} \in K\) for all \(k \in K\) but there is nothing allowing for us to extend this further and state that \(ghg^{-1} \in K\) for all \(g \in G\). \ However, a special case for when this is true involves \(Z(G)\). We know that \(Z(G) \normal G\). But if \(K \normal Z(G)\) then it turns out \(K \normal G\),