## Section15.4Normal Subgroups and Group Homomorphisms

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.

### Subsection15.4.1Normal Subgroups

We have seen that $$A_3= \left\{i,r_1,r_2\right\}$$ is a subgroup of $$S_3\text{,}$$ and its left cosets are $$A_3$$ itself and $$B_3=\left\{f_1,f_2,f_3\right\}\text{.}$$ 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. Figure 15.4.2. Operation table for $$S_3$$

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 $$A_3\circ B_3\text{.}$$ 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{equation*} \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} \end{equation*}

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

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

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 $$\left\{C_0,C_1,C_2\right\}\text{,}$$ 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:

\begin{equation*} a'*b' = \left(a*h_1\right)*\left(b*h_2 \right) = (a*b)*\left(h_1*h_2\right) \end{equation*}

since $$G$$ was abelian.

The last step was made possible by the fact that $$h_1*b=b*h_1\text{.}$$ 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 $$h_1*b = b*(\textrm{something in } H) = b*h_3\text{.}$$ 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.

#### Definition15.4.5.Normal Subgroup.

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

We leave the proof of this theorem to the reader.

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

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

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 $$\mathbb{Z}_2\text{.}$$

$$A_5\text{,}$$ 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, $$\sigma _1\sigma _2\cdots \sigma _k\text{,}$$ where the length of $$\sigma _i$$ is $$\ell_i\text{,}$$ then for any $$g\in S_n\text{,}$$ $$g^{-1}\circ f\circ g\text{,}$$ which is the conjugate of $$f$$ by $$g\text{,}$$ 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 $$g=(1,3,5,7,9)\text{,}$$

\begin{equation*} \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} \end{equation*}

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 $$A_5\text{.}$$ An outline of the process is in the exercises.

Let $$G$$ be the set of two by two invertible matrices of real numbers. That is,

\begin{equation*} 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\} \end{equation*}

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 $$H_2=\left\{\left.\left( \begin{array}{cc} a & 0 \\ 0 & d \\ \end{array} \right)\right| a d \neq 0\right\}\text{.}$$ It is fairly simple to apply one of the conditions we have observed for normallity that $$H_1$$ a normal subgroup of $$G\text{,}$$ while $$H_2$$ is not normal in $$G\text{.}$$

### Subsection15.4.2Homomorphisms

Think of the word isomorphism. Chances are, one of the first images that comes to mind is an equation something like

\begin{equation*} \theta(x * y) = \theta(x) \diamond \theta(y) \end{equation*}

An isomorphism must be a bijection, but the equation above is the algebraic property of an isomorphism. Here we will examine functions that satisfy equations of this type.

#### Definition15.4.12.Homomorphism.

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 $$x, y \in G\text{.}$$

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

Define $$\alpha:\mathbb{Z}_6\to \mathbb{Z}_3$$ by $$\alpha(n)=n \textrm{ mod } 3\text{.}$$ Therefore, $$\alpha(0) = 0\text{,}$$ $$\alpha(1) = 1\text{,}$$ $$\alpha(2) = 2\text{,}$$ $$\alpha(3) =1 + 1 + 1=0\text{,}$$ $$\alpha(4) = 1\text{,}$$ and $$\alpha(5) = 2\text{.}$$ If $$n, m \in \mathbb{Z}_6\text{.}$$ 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)\tag{15.4.1} \end{gather}

but we will use a line of reasoning that generalizes. We have already encountered Sun Tzu's Theorem, which implies that the function $$\beta: \mathbb{Z}_6\to \mathbb{Z}_3 \times \mathbb{Z}_2$$ defined by $$\beta(n)=(n\textrm{ mod } 3, n\textrm{ mod } 2)\text{.}$$ We need only observe that equating the first coordinates of both sides of the equation

\begin{gather} \beta(n +_6 m) = \beta(n)+\beta(m)\tag{15.4.2} \end{gather}

gives us precisely the homomorphism property.

1. Let $$a$$ be any element of $$G\text{.}$$ Then $$\theta(a) \in G'\text{.}$$

\begin{equation*} \begin{split} \theta(a)\diamond e' &= \theta(a) \quad \textrm{ by the definition of } e'\\ &=\theta(a*e)\quad \textrm{ by the definition of } e\\ &= \theta(a)\diamond \theta(e)\quad \textrm{ by the fact that } \theta \textrm{ is a homomorphism}\\ \end{split} \end{equation*}

By cancellation, $$e' = \theta(e)\text{.}$$

2. Again, let $$a \in G\text{.}$$ $$e' = \theta(e) = \theta\left(a*a^{-1} \right) = \theta(a)\diamond \theta\left(a^{-1}\right)\text{.}$$ Hence, by the uniqueness of inverses, $$\theta(a) ^{-1}= \theta\left(a^{-1}\right)\text{.}$$

3. Let $$b_1, b_2 \in \theta(H)\text{.}$$ Then there exists $$a_1, a_2\in H$$ such that $$\theta\left(a_1\right) = b_1\text{,}$$ $$\theta\left(a_2\right) = b_2\text{.}$$ Recall that a compact necessary and sufficient condition for $$H \leq G$$ is that $$x*y^{-1}\in H$$ for all $$x, y \in H\text{.}$$ Now we apply the same condition in $$G'\text{:}$$

\begin{equation*} \begin{split} b_1\diamond b_2{}^{-1} &= \theta\left(a_1\right)\diamond \theta\left(a_2\right){}^{-1}\\ & =\theta\left(a_1\right)\diamond \theta\left(a_2{}^{-1}\right)\\ & =\theta\left(a_1*a_2{}^{-1}\right)\in \theta(H)\\ \end{split} \end{equation*}

since $$a_1*a_2{}^{-1}\in H\text{,}$$ and so we can conclude that $$\theta(H)\leq G'\text{.}$$

If we define $$\pi: \mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z}$$ by $$\pi(n) = n + 4\mathbb{Z}\text{,}$$ then $$\pi$$ is a homomorphism. The image of the subgroup $$4\mathbb{Z}$$ is the single coset $$0 + 4\mathbb{Z}\text{,}$$ 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.

We leave the proof of this theorem to the reader.

#### Definition15.4.18.Natural Homomorphism.

If $$H \triangleleft G\text{,}$$ 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.

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

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

Let $$K=\textrm{ker }\theta\text{.}$$ We can see that $$K$$ is a subgroup of $$G$$ by letting $$a,b \in K$$ and verify that $$a*b^{-1} \in K$$ by computing $$\theta(a*b^{-1})= \theta(a)*\theta(b)^{-1} = e'*e'^{-1}=e'\text{.}$$ To prove normality, we let $$g$$ be any element of $$G$$ and $$k \in K\text{.}$$ We compute $$\theta(g*k*g^{-1})$$ to verify that $$g*k*g^{-1}\in K\text{.}$$

\begin{equation*} \begin{split} \theta(g*k*g^{-1}) &=\theta(g)*\theta(k)*\theta(g^{-1})\\ & =\theta(g)*\theta(k)*\theta(g)^{-1}\\ & =\theta(g)*e'*\theta(g)^{-1}\\ & =\theta(g)* \theta(g)^{-1}\\ & =e'\\ \end{split} \end{equation*}

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

Define $$\theta: \mathbb{Z} \rightarrow \mathbb{Z}_{10}$$ by $$\theta(n) = n \textrm{ mod }10\text{.}$$ 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.

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

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

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 $$\Phi(A) = \frac{A}{\sqrt{\lvert \det A \rvert }}\text{.}$$ We will let the reader verify that $$\Phi$$ is a homomorphism. The theorems above imply the following.

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

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

• $$\pi: G \rightarrow G\left/H_1\right.$$ 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.

Consider $$\Phi :\mathbb{Z}_2{}^2\to \mathbb{Z}_2{}^3$$ defined by $$\Phi (a, b) = \left(a, b, a +_2 b\right)\text{.}$$ If $$\left(a_1,b_1\right), \left(a _2 , b_2 \right) \in \mathbb{Z}_2{}^2\text{,}$$

\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 $$b = 0\text{,}$$ the kernel of $$\Phi$$ is $$\{(0, 0)\}\text{.}$$ 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 $$\mathbb{Z}_2{}^2\text{.}$$

We can generalize the previous example as follows: If $$n, m \geq 1$$ and $$A$$ is 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

\begin{equation*} \Phi \left(a_1, a_2 , . . . , a _m \right) = \left(a_1, a_2 , . . . , a _m\right)A \end{equation*}

is a homomorphism. This is true because matrix multiplication is distributive over addition. The only new idea here is that computation is done in $$\mathbb{Z}_2\text{.}$$ If $$a=\left(a_1, a_2 , . . . , a _m\right)$$ and $$b=\left(b_1, b_2 , . . . , b _m\right)\text{,}$$ $$(a + b)A = a A + b A$$ is true by basic matrix laws. Therefore, $$\Phi (a + b) = \Phi (a) + \Phi (b)\text{.}$$

### Exercises15.4.3Exercises

#### 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 $$\theta_1(a) =\left| a\right|\text{.}$$

2. $$\theta_2 : \mathbb{Z}_5 \rightarrow \mathbb{Z}_2$$ where $$\theta_2(n) =\left\{ \begin{array}{cc} 0 & \textrm{ if } n \textrm{ is even} \\ 1 & \textrm{ if } n \textrm{ is odd} \\ \end{array} \right.\text{.}$$

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

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

1. Yes, the kernel is $$\{1, -1\}$$

2. No, since $$\theta _2\left(2 +_{5} 4\right)= \theta_2(1)=1\text{,}$$ but $$\theta _2(2)+_2\theta_{2} (4)=0+_{2}0 =0$$

A follow-up might be to ask what happens if 5 is replaced with some other positive integer in this part.

3. Yes, the kernel is $$\{(a, -a)| a \in \mathbb{R}\}$$

4. No. A counterexample, among many, would be to consider the two transpositions $$t_1=(1,3)$$ and $$t_2=(1,2)\text{.}$$ Compare $$\theta_4(t_1 \circ t_2)$$ and $$\theta_4(t_1) \circ \theta_4(t_2)\text{.}$$

#### 2.

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

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

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

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

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

#### 3.

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

$$\langle r\rangle =\left\{i,r,r^2,r^3\right\}$$ is a normal subgroup of $$D_4\text{.}$$ To see you could use the table given in the solution of Exercise 15.3.5.5 of Section 15.3 and verify that $$a^{-1}h a \in \langle r\rangle$$ for all $$a\in D_4$$ and $$h\in \langle r\rangle\text{.}$$ A more efficient approach is to prove the general theorem that if $$H$$ is a subgroup $$G$$ with exactly two distinct left cosets, than $$H$$ is normal. $$\left\langle f_1\right\rangle$$ is not a normal subgroup of $$D_4\text{.}$$ $$\left\langle f_1\right\rangle =\left\{i,f_1\right\}$$ and if we choose $$a = r$$ and $$h=f_1$$ then $$a^{-1}h a= r^3f_1r=f_2\notin \left\langle f_1\right\rangle$$

#### 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 $$\alpha\left(a_1,a_2,a_3 \right) = \left(a_1,a_2,a_3 ,a_1+_2 a_2+_2a_3\right)\text{,}$$ and $$\beta \left(b_1,b_2,b_3,b_4\right)=b_1+b_2+b_3+b_4$$ Describe the function $$\beta \circ \alpha\text{.}$$ Is it a homomorphism?

$$(\beta \circ \alpha )\left(a_1,a_2,a_3\right) = 0$$ and so $$\beta \circ \alpha$$ is the trivial homomorphism, but a homomorphism nevertheless.

#### 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?

Let $$x, y \in G\text{.}$$

\begin{equation*} \begin{split} q(x * y) &= (x * y)^2\\ & = x*y*x*y \\ & = x*x*y*y \quad\textrm{ since } G \textrm{ is abelian}\\ & =x^2*y^2 \\ &= q(x)*q(y) \end{split} \end{equation*}

Hence, $$q$$ is a homomorphism. In order for $$q$$ to be an isomorphism, it must be the case that no element other than the identity is its own inverse.

\begin{equation*} \begin{split} x \in \textrm{Ker}(q) & \Leftrightarrow q (x) = e \\ & \Leftrightarrow x * x =e \\ & \Leftrightarrow x^{-1}= x\\ \end{split} \end{equation*}

#### 8.

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

#### 9.

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

Proof: Recall that the inverse image of $$H'$$ under $$\theta$$ is $$\theta ^{-1}(H')=\{g\in G | \theta (g)\in H'\}\text{.}$$

Closure: Let $$g_1, g_2\in \theta ^{-1}(H')\text{,}$$ then $$\theta \left(g_1\right),\theta \left(g_2\right)\in H'\text{.}$$ Since $$H'$$ is a subgroup of $$G'\text{,}$$

\begin{equation*} \theta\left(g_1\right)\diamond \theta\left(g_2\right)=\theta\left(g_1*g_2\right)\in H' \Rightarrow g_1*g_2\in \theta^{-1}(H') \end{equation*}

Identity: By Theorem 15.4.14(a), $$e \in \theta ^{-1}(H')\text{.}$$

Inverse: Let $$a\in \theta ^{-1}(H')$$ . Then $$\theta (a)\in H'$$ and by Theorem 15.4.14(b), $$\theta (a)^{-1}= \theta \left(a^{-1}\right)\in H'$$ and so $$a^{-1}\in \theta ^{-1}(H')\text{.}$$

#### 10.

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 $$\phi ^{-1}\circ f\circ \phi = g\text{.}$$

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.