# Section15.4Normal Subgroups and Group Homomorphisms¶ permalink

Our goal in this section is to answer an open question from earlier in this chapter and introduce a related concept. The question is: When are left cosets of a subgroup a group under the induced operation? This question is open for non-abelian groups. Now that we have some examples to work with, we can try a few experiments.

##### Example15.4.1Cosets of $A_3$

We have seen that $A_3= \left\{i,r_1,r_2\right\}$ is a subgroup of $S_3$, and its left cosets are $A_3$ itself and $B_3=\left\{f_1,f_2,f_3\right\}$. Whether $\left\{A_3 , B_3 \right\}$ is a group boils down to determining whether the induced operation is well defined. Consider the operation table for $S_3$ in Figure 15.4.2.

We have shaded in all occurrences of the elements of $B_3$ in gray. We will call these elements the gray elements and the elements of $A_3$ the white ones.

Now consider the process of computing the coset product $A_3\circ B_3$. The “product” is obtained by selecting one white element and one gray element. Note that white “times” gray is always gray. Thus, $A_3\circ B_3$ is well defined. Similarly, the other three possible products are well defined. The table for the factor group $S_3/A_3$ is $\begin{array}{c|c} \circ & \begin{array}{cc} A_3 & B_3 \\ \end{array} \\ \hline \begin{array}{c} A_{3 } \\ B_3 \\ \end{array} & \begin{array}{cc} A_3 & B_3 \\ B_3 & A_3 \\ \end{array} \\ \end{array}$

Clearly, $S_3/A_3$ is isomorphic to $\mathbb{Z}_2$. Notice that $A_3$ and $B_3$ are also the right cosets of $A_3$. This is significant.

##### Example15.4.3Cosets of another subgroup of $S_3$

Now let's try the left cosets of $\langle f_1 \rangle$ in $S_3$. There are three of them. Will we get a complicated version of $\mathbb{Z}_3$? The left cosets are $C_0=\left\langle f_1\right\rangle$, $C_1= r_1\left\langle f_1\right\rangle = \left\{r_1,f_3\right\}$, and $C_2= r_2\left\langle f_1\right\rangle = \left\{r_2,f_2\right\}$.

The reader might be expecting something to go wrong eventually, and here it is. To determine $C_1\circ C_2$ we can choose from four pairs of representatives: \begin{equation*} \begin{array}{c} r_1 \in C_1, r_2\in C_2 \longrightarrow r_1\circ r_2=i\in C_0\\ r_1\in C_1, f_2\in C_2 \longrightarrow r_1\circ f_2=f\in C_0\\ f_3\in C_1, r_2\in C_2 \longrightarrow f_3\circ r_2=f_2\in C_2\\ f_3\in C_1, f_2\in C_2 \longrightarrow f_3\circ f_2=r_2\in C_2\\ \end{array} \end{equation*} This time, we don't get the same coset for each pair of representatives. Therefore, the induced operation is not well defined and no factor group is produced.

##### Observation15.4.4

This last example changes our course of action. If we had gotten a factor group from $\left\{C_0,C_1,C_2\right\}$, we might have hoped to prove that every collection of left cosets forms a group. Now our question is: How can we determine whether we will get a factor group? Of course, this question is equivalent to: When is the induced operation well defined? There was only one step in the proof of Theorem 15.2.18, where we used the fact that $G$ was abelian. We repeat the equations here: $a'*b' = \left(a*h_1\right)*\left(b*h_2 \right) = (a*b)*\left(h_1*h_2\right)$ since $G$ was abelian.

The last step was made possible by the fact that $h_1*b=b*h_1$. As the proof continued, we used the fact that $h_1*h_2$ was in $H$ and so $a'*b'$ is $(a*b)*h$ for some $h$ in $H\text{.}$ All that we really needed in the “abelian step” was that $h_1*b = b*(\textrm{ something in } H) = b*h_3$. Then, since $H$ is closed under $G$'s operation, $h_3*h_2$ is an element of $H\text{.}$ The consequence of this observation is that we define a certain kind of subgroup that guarantees that the inducted operation is well-defined.

##### Definition15.4.5Normal Subgroup.

If $G$ is a group, $H \leq G$, then $H$ is a normal subgroup of $G$, denoted $H \triangleleft G$, if and only if every left coset of $H$ is a right coset of $H$; i. e. $a*H = H*a \quad \forall a \in G$

##### Proof

Be careful, the following corollary is not an “...if and only if...” statement.

##### Example15.4.8A non-normal subgroup

The right cosets of $\left\langle f_1\right\rangle \leq S_3$ are $\left\{i, f_1\right\}$, $\left\{r_1 f_2 \right\}$, and $\left\{r_2 ,f_3\right\}$. These are not the same as the left cosets of $\left\langle f_1\right\rangle$. In addition, $f_2{}^{-1}f_1f_2=f_2f_1f_2=f_3\notin \left\langle f_1\right\rangle$. Thus, $\left\langle f_1\right\rangle$ is not normal.

The improper subgroups $\{e\}$ and $G$ of any group $G$ are normal subgroups. $G/\{e\}$ is isomorphic to $G\text{.}$ All other normal subgroups of a group, if they exist, are called proper normal subgroups.

##### Example15.4.9

By Condition b of Corollary 15.4.7, $A_n$ is a normal subgroup of $S_n$ and $S_n/A_n$ is isomorphic to $\mathbb{Z}_2$.

##### Example15.4.10Subgroups of $A_5$

$A_5$, a group in its own right with 60 elements, has many proper subgroups, but none are normal. Although this could be done by brute force, the number of elements in the group would make the process tedious. A far more elegant way to approach the verification of this statement is to use the following fact about the cycle structure of permutations. If $f\in S_n$ is a permutation with a certain cycle structure, $\sigma _1\sigma _2\cdots \sigma _k$, where the length of $\sigma _i$ is $\ell_i$, then for any $g\in S_n$, $g^{-1}\circ f\circ g$, which is the conjugate of f by g, will have a cycle structure with exactly the same cycle lengths. For example if we take $f=(1,2,3,4)(5,6)(7,8,9)\in S_9$ and conjugate by $g=(1,3,5,7,9)$, $\begin{split} g^{-1}\circ f\circ g & =(1,9,7,5,3)\circ (1,2,3,4)(5,6)(7,8,9)\circ (1,3,5,7,9)\\ & = (1,4,9,2)(3,6)(5,8,7)\\ \end{split}$

Notice that the condition for normality of a subgroup $H$ of $G$ is that the conjugate of any element of $H$ by an element of $G$ must be remain in $H\text{.}$

To verify that $A_5$ has no proper normal subgroups, you can start by cataloging the different cycle structures that occur in $A_5$ and how many elements have those structures. Then consider what happens when you conjugate these different cycle structures with elements of $A_5$. An outline of the process is in the exercises.

##### Example15.4.11

Let $G$ be the set of two by two invertible matrices of real numbers. That is, $G=\left\{\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) \mid a,b,c,d\in \mathbb{R}, a d-b c\neq 0\right\}$ We saw in Chapter 11 that $G$ is a group with matrix multiplication.

This group has many subgroups, but consider just two: $H_1=\left\{\left.\left( \begin{array}{cc} a & 0 \\ 0 & a \\ \end{array} \right)\right| a \neq 0\right\}$ and $H_2=\left\{\left.\left( \begin{array}{cc} a & 0 \\ 0 & d \\ \end{array} \right)\right| a d \neq 0\right\}$. It is fairly simple to apply one of the conditions we have observed for normallity that $H_1$ a normal subgroup of $G$, while $H_2$ is not normal in $G$.

Think of the word isomorphism. Chances are, one of the first images that comes to mind is an equation something like $\theta(x * y) = \theta(x) \diamond \theta(y)$ An isomorphism must be a bijection, but equation the equation above is the algebraic feature of an isomorphism. Here we will examine functions that satisfy equations of this type.

##### Definition15.4.12Homomorphism

Let $[G, *]$ and $[G',\diamond ]$ be groups. $\theta:G \to G'$ is a homomorphism if $\theta(x * y) = \theta(x) \diamond \theta(y)$ for all $x, y \in G$.

Many homomorphisms are useful since they point out similarities between the two groups (or, on the universal level, two algebraic systems) involved.

##### Example15.4.13Decreasing modularity

Define $\alpha:\mathbb{Z}_6\to \mathbb{Z}_3$ by $\alpha(n)=n mod 3$. Therefore, $\alpha(0) = 0$, $\alpha(1) = 1$, $\alpha(2) = 2$, $\alpha(3) =1 + 1 + 1=0$, $\alpha(4) = 1$, and $\alpha(5) = 2$. If $n, m \in \mathbb{Z}_6$. We could actually show that $\alpha$ is a homomorphism by checking all $6^2=36$ different cases for the formula \begin{gather} \alpha(n +_6 m) = \alpha(n)+_3\alpha(m)\label{ex-15-4-h2}\tag{15.4.1} \end{gather} but we will use a line of reasoning that generalizes. We have already encountered the Chinese Remainder Theorem, which implies that the function $\beta: \mathbb{Z}_6\to \mathbb{Z}_3 \times \mathbb{Z}_2$ defined by $\beta(n)=(n\textrm{ mod } 3, n\textrm{ mod } 2)$. We need only observe that equating the first coordinates of both sides of the equation \begin{gather} \beta(n +_6 m) = \beta(n)+_3\beta(m)\label{mrow-64}\tag{15.4.2} \end{gather} gives us precisely the homomorphism property.

##### Example15.4.16

If we define $\pi: \mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z}$ by $\pi(n) = n + 4\mathbb{Z}$. then $\pi$ is a homomorphism. The image of the subgroup $4\mathbb{Z}$ is the single coset $0 + 4\mathbb{Z}$, the identity of the factor group. Homomorphisms of this type are called natural homomorphisms. The following theorems will verify that $\pi$ is a homomorphism and also show the connection between homomorphisms and normal subgroups. The reader can find more detail and proofs in most abstract algebra texts.

##### Definition15.4.18Natural Homomorphism

If $H \triangleleft G$, then the function $\pi:G\to G/H$ defined by $\pi(a) = a H$ is called the natural homomorphism.

Based on Theorem 15.4.17, every normal subgroup gives us a homomorphism. Next, we see that the converse is true.

##### Definition15.4.19Kernel of a homomorphism

Let $\theta: G \to G'$ be a homomorphism, and let $e$ and $e'$ be the identitites of $G$ and $G'$, respectively. The kernel of $\theta$ is the set $\ker \theta=\{a\in G \mid \theta(a)=e'\}$

##### Proof

Based on this most recent theorem, every homomorphism gives us a normal subgroup.

##### Example15.4.22

Define $\theta: \mathbb{Z} \rightarrow \mathbb{Z}_{10}$ by $\theta(n) = n mod 10$. The three previous theorems imply the following:

• $\pi: \mathbb{Z} \rightarrow \mathbb{Z}/10\mathbb{Z}$ defined by $\pi(n) = n + 10\mathbb{Z}$ is a homomorphism.

• $\{n\in \mathbb{Z}|\theta(n) = 0\} = \{10n \mid n \in \mathbb{Z}\}= 10\mathbb{Z} \triangleleft \mathbb{Z}$.

• $\mathbb{Z}/10\mathbb{Z}$ is isomorphic to $\mathbb{Z}_{10}$.

##### Example15.4.23

Let $G$ be the same group of two by two invertible real matrices as in Example 15.4.11. Define $\Phi: G \rightarrow G$ by $\Phi(A) = \frac{A}{\sqrt{\lvert \det A \rvert }}$. We will let the reader verify that $\Phi$ is a homomorphism. The theorems above imply the following.

• $\ker \Phi = \{A\in G |\Phi (A) =I\} = \left\{\left( \begin{array}{cc} a & 0 \\ 0 & a \\ \end{array} \right) \mid a\in \mathbb{R},a\neq 0\right\}\triangleleft G$. This verifies our statement in Example 15.4.11. As in that example, let $\ker \Phi = H_1$.

• $G\left/H_1\right.$ is isomorphic to $\{A \in G \mid \det A= 1\}$.

• $\pi: G \rightarrow G\left/H_1\right.$ defined, naturally, by $\pi(A) =A H_1$ is a homomorphism.

For the remainder of this section, we will be examining certain kinds of homomorphisms that will play a part in our major application to homomorphisms, coding theory.

##### Example15.4.24

Consider $\Phi :\mathbb{Z}_2{}^2\to \mathbb{Z}_2{}^3$ defined by $\Phi (a, b) = \left(a, b, a +_2 b\right)$. If $\left(a_1,b_1\right), \left(a _2 , b_2 \right) \in \mathbb{Z}_2{}^2$, \begin{equation*} \begin{split} \Phi\left(\left(a_1,b_1\right)+\left(a _2 , b_2 \right)\right) &= \Phi\left(a_1+_2a_2,b_1 +_2 b_2 \right)\\ & = \left(a_1+_2a_2,b_1 +_2 b_2 ,a_1+_2a_2+_2b_1 +_2 b_2\right)\\ & = \left(a_1,b_1 , a_1+_2b_1\right)+\left(a_2,b_2 , a_2+_2b_2\right)\\ & =\Phi \left(a_1,b_1\right)+\Phi \left(a _2 , b_2 \right)\\ \end{split} \end{equation*}

Since $\Phi (a, b)\text = (0, 0, 0)$ implies that $a = 0$ and $b = 0$, the kernel of $\Phi$ is $\{(0, 0)\}$. By previous theorems, $\Phi \left(\mathbb{Z}_2{}^2\right)= \{(0, 0, 0), (1, 0, 1), (0, 1, 1), (1, 1, 0)\}$ is isomorphic to $\mathbb{Z}_2{}^2$.

We can generalize the previous example as follows: If $n, m \geq 1$ and $A$ an $m\times n$ matrix of 0's and 1's (elements of $\mathbb{Z}_2$), then $\Phi :\mathbb{Z}_2{}^m\to \mathbb{Z}_2{}^n$ defined by $\Phi \left(a_1, a_2 , . . . , a _m \right) = \left(a_1, a_2 , . . . , a _m\right)A$ is a homomorphism. This is true because matrix multiplication is distributive over addition. The only new idea here is that computation is done in $\mathbb{Z}_2$. If $a=\left(a_1, a_2 , . . . , a _m\right)$ and $b=\left(b_1, b_2 , . . . , b _m\right)$, $(a + b)A = a A + b A$ is true by basic matrix laws. Therefore, $\Phi (a + b) = \Phi (a) + \Phi (b)$.

# Subsection15.4.3Exercises for Section 15.4¶ permalink

##### 1

Which of the following functions are homomorphisms? What are the kernels of those functions that are homomorphisms?

1. $\theta_1: \mathbb{R}^* \to \mathbb{R}^+$ defined by $\theta_1(a) =\left| a\right|$.

2. $\theta_2 : \mathbb{Z}_8 \rightarrow \mathbb{Z}_2$ where $\theta_2(n) =\left\{ \begin{array}{cc} 0 & \textrm{ if} n \textrm{ is} \textrm{ even} \\ 1 & \textrm{ if} n \textrm{ is} \textrm{ odd} \\ \end{array} \right.$ .

3. $\theta_3 : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, where $\theta_3(a, b) = a + b$.

4. $\theta_4 : S_4 \to S_4$ defined by $\theta_4(f) = f\circ f=f^2$.

##### 2

Which of the following functions are homomorphisms? What are the kernels of those functions that are homomorphisms?

1. $\alpha_1: M_{2\times 2}(\mathbb{R}) \rightarrow \mathbb{R}$, defined by $\alpha_1(A) = A_{11} A_{22} + A_{12} A_{21\textrm{ \ss}}$.

2. $\alpha_2 : \left(\mathbb{R}^*\right)^2 \rightarrow \mathbb{R}^*$ defined by $\alpha_2 (a, b) = a b$.

3. $\alpha_3 : \left\{\left.A \in M_{2\times 2}(\mathbb{R}) \right| \det A \neq 0\right\} \to \mathbb{R}^*$, where $\alpha_3(A) = \det A$.

4. $\alpha_4 : S_4\rightarrow S_4$ defined by $\alpha_4(f)=f^{-1}$.

##### 3

Show that $D_4$ has one proper normal subgroup, but that $\langle (1,4)(2,3)\rangle$ is not normal.

##### 4

Prove that the function $\Phi$ in Example 15.4.23 is a homomorphism.

##### 5

Define the two functions $\alpha: \mathbb{Z}_2{}^3\rightarrow \mathbb{Z}_2{}^4$ and $\beta :\mathbb{Z}_2{}^4\to \mathbb{Z}_2$ by $\alpha\left(a_1,a_2,a_3 \right) = \left(a_1,a_2,a_3 ,a_1+_2 a_2+_2a_3\right)$, and $\beta \left(b_1,b_2,b_3,b_4\right)=b_1+b_2+b_3+b_4$ Describe the function $\beta \circ \alpha$. Is it a homomorphism?

##### 7

Prove that if $G$ is an abelian group, then $q(x) = x^2$ defines a homomorphism from $G$ into $G\text{.}$ Is $q$ ever an isomorphism?

##### 8

Prove that if $\theta : G\rightarrow G'$ is a homomorphism, and $H\triangleleft G$, then $\theta(H) \triangleleft \theta(G)$. Is it also true that $\theta(H) \triangleleft G'$?

##### 9

Prove that if $\theta : G \rightarrow G'$ is a homomorphism, and $H' \leq \theta(G)$, then $\theta^{-1}(H') =\{a\in G| \theta (a)\in H'\}\leq G$.

Following up on Example 15.4.10, prove that $A_5$ is a simple group; i. e., it has no proper normal subgroups.
1. Make a list of the different cycle structures that occur in $A_5$ and how many elements have those structures.
2. Within each set of permutations with different cycle structures, identify which subsets are closed with respect to the conjugation operation. With this you will have a partition of $A_5$ into conjugate classes. where for each class $C\text{,}$ $f, g\in C$ if and only if $\exists \phi \in A_5$ such that $\phi ^{-1}\circ f\circ \phi = g$
3. Use the fact that a normal subgroup of $A_5$ needs to be a union of conjugate classes and verify that no such union exists.