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Section 3.2 Factor Groups and Normal Subgroups

If \(H\) is a subgroup of a group \(G\text{,}\) then right cosets are not always the same as left cosets; that is, it is not always the case that \(gH = Hg\) for all \(g \in G\text{.}\) The subgroups for which this property holds play a critical role in group theory— they allow for the construction of a new class of groups, called factor or quotient groups. Factor groups may be studied directly or by using homomorphisms, a generalization of isomorphisms. We will study isomorphisms and homomorphisms in Chapter 4.

Subsection 3.2.1 Normal Subgroups

A subgroup \(H\) of a group \(G\) is normal in G if \(gH = Hg\) for all \(g \in G\text{.}\) That is, a normal subgroup of a group \(G\) is one in which the right and left cosets are precisely the same.

Example 3.9.

Let \(G\) be an abelian group. Every subgroup \(H\) of \(G\) is a normal subgroup. Since \(gh = hg\) for all \(g \in G\) and \(h \in H\text{,}\) it will always be the case that \(gH = Hg\text{.}\)

Example 3.10.

Let \(H\) be the subgroup of \(S_3\) consisting of elements \((1)\) and \((12)\text{.}\) Since

\begin{equation*} (123) H = \{ (123), (13) \} \quad \text{and} \quad H (123) = \{ (123), (23) \}\text{,} \end{equation*}

\(H\) cannot be a normal subgroup of \(S_3\text{.}\) However, the subgroup \(N\text{,}\) consisting of the permutations \((1)\text{,}\) \((123)\text{,}\) and \((132)\text{,}\) is normal since the cosets of \(N\) are

\begin{gather*} N = \{ (1), (123), (132) \}\\ (12) N = N (12) = \{ (12), (13), (23) \}\text{.} \end{gather*}

The following theorem is fundamental to our understanding of normal subgroups.

Proof.

(1) \(\Rightarrow\) (2). Since \(N\) is normal in \(G\text{,}\) \(gN = Ng\) for all \(g \in G\text{.}\) Hence, for a given \(g \in G\) and \(n \in N\text{,}\) there exists an \(n'\) in \(N\) such that \(g n = n' g\text{.}\) Therefore, \(gng^{-1} = n' \in N\) or \(gNg^{-1} \subset N\text{.}\)

(2) \(\Rightarrow\) (3). Let \(g \in G\text{.}\) Since \(gNg^{-1} \subset N\text{,}\) we need only show \(N \subset gNg^{-1}\text{.}\) For \(n \in N\text{,}\) \(g^{-1}ng=g^{-1}n(g^{-1})^{-1} \in N\text{.}\) Hence, \(g^{-1}ng = n'\) for some \(n' \in N\text{.}\) Therefore, \(n = g n' g^{-1}\) is in \(g N g^{-1}\text{.}\)

(3) \(\Rightarrow\) (1). Suppose that \(gNg^{-1} = N\) for all \(g \in G\text{.}\) Then for any \(n \in N\) there exists an \(n' \in N\) such that \(gng^{-1} = n'\text{.}\) Consequently, \(gn = n' g\) or \(gN \subset Ng\text{.}\) Similarly, \(Ng \subset gN\text{.}\)

Subsection 3.2.2 Factor Groups

If \(N\) is a normal subgroup of a group \(G\text{,}\) then the cosets of \(N\) in \(G\) form a group \(G/N\) under the operation \((aN) (bN) = abN\text{.}\) This group is called the factor or quotient group of \(G\) and \(N\text{.}\) Our first task is to prove that \(G/N\) is indeed a group.

Proof.

The group operation on \(G/N\) is \((a N ) (b N)= a b N\text{.}\) This operation must be shown to be well-defined; that is, group multiplication must be independent of the choice of coset representative. Let \(aN = bN\) and \(cN = dN\text{.}\) We must show that

\begin{equation*} (aN) (cN) = acN = bd N = (b N)(d N)\text{.} \end{equation*}

Then \(a = b n_1\) and \(c = d n_2\) for some \(n_1\) and \(n_2\) in \(N\text{.}\) Hence,

\begin{align*} acN & = b n_1 d n_2 N\\ & = b n_1 d N\\ & = b n_1 N d\\ & = b N d\\ & = b d N\text{.} \end{align*}

The remainder of the theorem is easy: \(eN = N\) is the identity and \(g^{-1} N\) is the inverse of \(gN\text{.}\) The order of \(G/N\) is, of course, the number of cosets of \(N\) in \(G\text{.}\)

It is very important to remember that the elements in a factor group are sets of elements in the original group.

Example 3.13.

Consider the normal subgroup of \(S_3\text{,}\) \(N = \{ (1), (123), (132) \}\text{.}\) The cosets of \(N\) in \(S_3\) are \(N\) and \((12) N\text{.}\) The factor group \(S_3 / N\) has the following multiplication table.

\begin{equation*} \begin{array}{c|cc} & N & (12) N \\ \hline N & N & (12) N \\ (12) N & (12) N & N \end{array} \end{equation*}

This group has order 2, and you might think it looks a lot like \({\mathbb Z}_2\) (in Chapter 4 we will see that this group is isomorphic to \(\Z_2\)). At first, multiplying cosets seems both complicated and strange; however, notice that \(S_3 / N\) is a smaller group. The factor group displays a certain amount of information about \(S_3\text{.}\) Actually, \(N = A_3\text{,}\) the group of even permutations, and \((12) N = \{ (12), (13), (23) \}\) is the set of odd permutations. The information captured in \(G/N\) is parity; that is, multiplying two even or two odd permutations results in an even permutation, whereas multiplying an odd permutation by an even permutation yields an odd permutation.

Example 3.14.

Consider the normal subgroup \(3 {\mathbb Z}\) of \({\mathbb Z}\text{.}\) The cosets of \(3 {\mathbb Z}\) in \({\mathbb Z}\) are

\begin{align*} 0 + 3 {\mathbb Z} & = \{ \ldots, -3, 0, 3, 6, \ldots \}\\ 1 + 3 {\mathbb Z} & = \{ \ldots, -2, 1, 4, 7, \ldots \}\\ 2 + 3 {\mathbb Z} & = \{ \ldots, -1, 2, 5, 8, \ldots \}\text{.} \end{align*}

The group \({\mathbb Z}/ 3 {\mathbb Z}\) is given by the Cayley table below.

\begin{equation*} \begin{array}{c|ccc} + & 0 + 3{\mathbb Z} & 1 + 3{\mathbb Z} & 2 + 3{\mathbb Z} \\\hline 0 + 3{\mathbb Z} & 0 + 3{\mathbb Z} & 1 + 3{\mathbb Z} & 2 + 3{\mathbb Z} \\ 1 + 3{\mathbb Z} & 1 + 3{\mathbb Z} & 2 + 3{\mathbb Z} & 0 + 3{\mathbb Z} \\ 2 + 3{\mathbb Z} & 2 + 3{\mathbb Z} & 0 + 3{\mathbb Z} & 1 + 3{\mathbb Z} \end{array} \end{equation*}

In general, the subgroup \(n {\mathbb Z}\) of \({\mathbb Z}\) is normal. The cosets of \({\mathbb Z } / n {\mathbb Z}\) are

\begin{gather*} n {\mathbb Z}\\ 1 + n {\mathbb Z}\\ 2 + n {\mathbb Z}\\ \vdots\\ (n-1) + n {\mathbb Z}\text{.} \end{gather*}

The sum of the cosets \(k + n{\mathbb Z}\) and \(l + n{\mathbb Z}\) is \(k+l + n{\mathbb Z}\text{.}\) Notice that we have written our cosets additively, because the group operation is integer addition.

Example 3.15.

Consider the dihedral group \(D_n\text{,}\) generated by the two elements \(r\) and \(s\text{,}\) satisfying the relations

\begin{align*} r^n & = \identity\\ s^2 & = \identity\\ srs & = r^{-1}\text{.} \end{align*}

The element \(r\) actually generates the cyclic subgroup of rotations, \(R_n\text{,}\) of \(D_n\text{.}\) Since \(srs^{-1} = srs = r^{-1} \in R_n\text{,}\) the group of rotations is a normal subgroup of \(D_n\text{;}\) therefore, \(D_n / R_n\) is a group. The group has exactly two elements, so it will look just like \({\mathbb Z}_2\text{.}\)

Reading Questions 3.2.3 Reading Questions

1.

Give two ways we can can be sure that a subgroup \(H\) of a group \(G\) is normal.

2.

What does the notation \(G/H\) represent? What does the notation \([G:H]\) represent? How are these related?

3.

The subgroup \(8\Z\) is normal in \(\Z\text{.}\) In the factor group \(\Z/8\Z\text{,}\) perform the computation \((3+8\Z) + (7+8\Z)\text{.}\)

4.

After reading the section, what questions do you still have? Write at least one well formulated question (even if you think you understand everything).

Exercises 3.2.4 Practice Problems

1.

Let \(G = D_4\text{,}\) the group of symmetries of a square, and take \(H = \{r_0, r_2, d_1, d_2\}\text{.}\) Is \(H\) normal in \(G\text{?}\) If so, how many elements are in the factor group \(G/H\text{?}\) Write out its group table.

2.

A conjugate of an element \(x \in G\) is an element of \(G\) with the form \(gxg\inv\) for some \(g \in G\text{.}\) Using \(G = D_4\) and \(H = \{r_0, r_2, d_1, d_2\}\text{,}\) find all the conjugates of all the elements of \(H\text{.}\) What do you notice?

3.

Prove that if \(G\) is abelian and \(H\) is a subgroup, that \(G/H\) is abelian.

4.

True or false: If \(H\) is abelian and \(G/H\) is abelian, then \(G\) is abelian. Prove your answer.

Hint.

Look at Exercise 3.2.4.1 .

5.

Let \(G = D_4\) and \(H = \{r_0, h\}\text{.}\) Prove that \(H\) is not normal in \(G\text{.}\) Then give a specific example of what goes wrong if you try to define coset multiplication on the set of cosets \(G/H\text{.}\)

6.

Again let \(G = D_4\) and \(H = \{r_0, h\}\text{.}\) Show that \(H \) is not closed under conjugates.

Exercises 3.2.5 Collected Homework

C1.

Let \(G = S_3\) and \(H = \{(1), (13)\}\text{.}\)

  1. Find the three (left) cosets of \(H\) in \(G\text{.}\)

  2. Explain what goes wrong if you try to define an operation on the set of cosets \(G/H\text{.}\) Do this with a specific example using this \(G\) and \(H\text{.}\)

  3. Write out all the conjugates of \(H\) in \(G\text{.}\) That is, find \(gxg\inv\) for all \(x \in H\) and \(g \in G\text{.}\) Is \(H\) closed under conjugates?

  4. Show that \(H\) is not normal in \(G\) using the definition (about left and right cosets).

C2.

BONUS! Prove that if \([G : H] = 2\) then \(H\) is normal in \(G\text{.}\)

Hint.

This is not all that hard. Try an example (such as Exercise 3.2.4.1 ) to see why this works. Then explain it in general.