12.2 Normal Subgroups and Factor Groups

12.2.1 Cosets

Definition 12.3 Let \(S\) be a subgroup of \(G\). The subset \(aS = \left\{ a b: b\in S \right\}\) of \(G\) is the left coset of \(S\) containing \(a\) and \(Sa = \left\{ ba : b\in S \right\}\) is the right coset of \(S\) containing \(a\).

Theorem 12.2 For a group \(G\) with identity \(e\), for each element \(a\in G\) + \(aG=Ga=G\) + \(a\{e\}=\{e\} a=\{a\}\)

Proof. In order to prove that the left and right cosets of \(G\) containing \(a\) are always the entire group \(G\), we choose a generic element of \(G\) and we can call it \(b\). Since \(G\) is a group, \(a^{-1}b\in G\) and so \(b= a (a^{-1}b) \in aG\). Similarly, since \(ba^{-1}\in G\), \(b = (ba^{-1})a \in Ga\). So \(aG=Ga=G\).

Since \(ae=a\), we can also see that \(a\{e\} = \{a\}\) and \(\{e\}a=\{a\}\).

If we return to the group \(D_3\), we can look at the left and right cosets related to the corresponding non-trivial subgroups. If we let \(S=\{r_0,s_1\}\), we have the following left and right cosets,

\[\begin{array}{|c|c|} \hline \mbox{Left Cosets of } \{r_0, s_1\} & \mbox{Right Cosets of } \{r_0, s_1\}\\ \hline r_0 \circ S = \{r_0, s_1\} = S & S \circ r_0 = \{r_0, s_1\} = S \\ r_1 \circ S = \{r_1, s_3\} & S \circ r_1 = \{r_1, s_2\} \\ r_2 \circ S = \{r_2, s_2 \} & S \circ r_2 = \{r_2, s_3 \} \\ s_1 \circ S = \{s_1, r_0\}=S & S \circ s_1 = \{s_1, r_0\} = S \\ s_2 \circ S = \{s_2, r_2 \} & S \circ s_2 = \{s_2, r_1\} \\ s_3 \circ S = \{s_3, r_1\} & S \circ s_3 = \{s_3, r_2\} \\ \hline \end{array}\]

and we see that many of the left cosets are different from the corresponding right cosets. On the other hand, if we let \(S=\{r_0, r_1, r_2\}\) we have the following left and right cosets.

\[\begin{array}{|c|c|} \hline \mbox{Left Cosets of } \{r_0, r_1, r_2\} & \mbox{Right Cosets of } \{r_0, r_1, r_2\}\\ \hline r_0 \circ S = \{r_0, r_1, r_2\} = S & S \circ r_0 = \{r_0, r_1, r_2\} = S \\ r_1 \circ S = \{r_1, r_2, r_3 \} = S & S \circ r_1 = \{r_1, r_2, r_3\} =S \\ r_2 \circ S = \{r_2, r_0, r_1 \}=S & S \circ r_2 = \{r_2, r_0, r_1 \}=S \\ s_1 \circ S = \{s_1, s_2, s_3\} & S \circ s_1 = \{s_1, s_3, s_2\} \\ s_2 \circ S = \{s_2, s_3, s_1 \} & S \circ s_2 = \{s_2, s_1, s_3\} \\ s_3 \circ S = \{s_3, s_1, s_2 \} & S \circ s_3 = \{s_3, s_2, s_1 \} \\ \hline \end{array}\]

In this situation, we see that the left cosets and the right cosets correspond and that the group \(D_3\) is partitioned into two sets, \(\{r_0, r_1, r_2\}\) and \(\{s_1, s_2, s_3\}\) by the two cosets.

12.2.2 Normal Subgroups

The subgroups for which the left and right cosets correspond are called normal.

Definition 12.4 A subgroup \(N\) of a group \(G\) is called a normal subgroup if it is invariant under conjugation, that is if \(gN = Ng\) for all \(g\in G\). We denote this by \(N \vartriangleleft G\).

Theorem 12.3 A subgroup \(N\) of a group \(G\) is a normal subgroup if and only if for all \(g\in G\) and \(f\in N\), \(g^{-1}*f*g \in N\).

Proof. Let’s first assume that \(N\) is a normal subgroup of \(G\). Then for all \(g\in G\), \(gN=Ng\). If \(f\in N\), we see that \(fg\in gN\) and so there exists \(\tilde{f}\in N\) such that \(fg=g\tilde{f}\) and so \(g^{-1}fg \in N\). Since \(f\) and \(g\) were arbitrary, \(g^{-1}*f*g \in N\) for all \(g\in G\) and \(f\in N\).

Let’s alternatively assume that \(g^{-1}*f*g \in N\) for all \(g\in G\) and \(f\in N\). Then if we choose an element \(g\in G\), we know that \(gN = \left\{ g f \: \vert \: f\in N\right\}\) and \(Ng=\left\{ fg \: \vert \: f\in N\right\}\). If \(fg\in Ng\), we know that \(g^{-1} fg \in N\) and so there exists \(\tilde{f}\in N\) such that \(g^{-1}fg=\tilde{f}\) and \(fg=g\tilde{f}\), implying that \(fg\in gN\). So \(Ng\subseteq gN\). We can similarly show that \(gN\subseteq Ng\) and so \(gN=Ng\) and \(N\) is subnormal.

Theorem 12.4 If \(N\) is a normal subgroup of \(G\), then for every \(n\in N\), \(nN=N\).

12.2.3 Factor Groups

If \(N\) is a normal subgroup of a group \(G\), we can create a relation on \(G\) as follows: For \(g,h\in G\) \[g \sim h \Leftrightarrow gN=hN.\]

Since \(gN=gN\), we see that the relation is reflexive. Since \(gN=hN\) is equivalent to \(hN=gN\) based on properties of sets we see that the relation is symmetric. In order to show that the relation is transitive, we assume that \(g,h,m\in G\) such that \(g \sim h\) and \(h\sim m\) and see that transitivity of set equality implies that \(g\sim m\) and so the relation is transitive.

Since this relation is an equivalence relation we see that it creates a partition of \(G\) into cosets. We can then define \[\frac{G}{N} = \left\{ gN \: \vert \: g\in G\right\}.\] We can also define an operation on this set using the operation from \(G\) as \[(gN)(hN) := (gh)N.\] To verify that this operation is well defined, we can choose two other representations of these cosets so that \(gN=g_1 N\) and \(hN=h_1N\) and verify that the product of the cosets is the same independent of the representation. Since \(gN=g_1N\), there exists \(n_1\in N\) such that \(gn_1 =g_1\), and since \(hN=h_1N\), there exists \(n_2\in N\) such that \(hn_2=h_1\). Therefore, \[(g_1 h_1)N= (g n_1)(h n_2) N = g (n_1 h)N\] since \(n_2N=N\). And since \(N\) is normal, there exist \(n_3\in N\) such that \(n_1 h= h n_3\), so \[(g_1 h_1)N= (gh)n_1 N = (gh)N.\] So the operation is well-defined.

Theorem 12.5 If \(N\) is a normal subgroup of \(G\), then \(\frac{G}{N}\) with the operation inherited from \(G\) is a group.

This group \(\frac{G}{N}\) is called the quotient group of \(G\) with respect to \(N\).

12.2.4 Exercises

  1. Consider the relationship between \((\mathbb{Z},+)\) and \((\mathbb{R},+)\).

    1. Show that \((\mathbb{Z},+)\) is a normal subgroup of \((\mathbb{R},+)\).
    2. Describe the elements of the factor group \[\frac{(\mathbb{R},+)}{(\mathbb{Z},+)}\] and the operation induced from \((\mathbb{R},+)\).

  2. If a group is abelian, what does this tell us about normality of the subgroups? Why?

  3. Prove that the intersection of normal subgroups of a group \(G\) is again a normal subgroup of \(G\).