ads Normal Subgroups and Group Homomorphisms

Section 15.4 Normal 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.

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Subsection 15.4.1 Normal Subgroups

Example 15.4.1. Cosets of $A_{3}$ .

We have seen that

A_{3} = {i, r_{1}, r_{2}}

is a subgroup of

S_{3},

and its left cosets are

A_{3}

itself and

B_{3} = {f_{1}, f_{2}, f_{3}} .

Whether

{A_{3}, B_{3}}

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.

described in detail following the image — 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} .

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}

\begin{array}{cc} \circ & \begin{array}{cc} A_{3} & B_{3} \end{array} \\ \begin{matrix} A_{3} \\ B_{3} \end{matrix} & \begin{array}{cc} A_{3} & B_{3} \\ B_{3} & A_{3} \end{array} \end{array}

Clearly,

S_{3} / A_{3}

is isomorphic to

Z_{2} .

Notice that

A_{3}

and

B_{3}

are also the right cosets of

A_{3} .

This is significant.

Example 15.4.3. Cosets of another subgroup of $S_{3}$ .

Now let’s try the left cosets of

⟨ f_{1} ⟩

S_{3} .

There are three of them. Will we get a complicated version of

Z_{3} ?

The left cosets are

C_{0} = ⟨ f_{1} ⟩,

C_{1} = r_{1} ⟨ f_{1} ⟩ = {r_{1}, f_{3}},

and

C_{2} = r_{2} ⟨ f_{1} ⟩ = {r_{2}, f_{2}} .

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{matrix} r_{1} \in C_{1}, r_{2} \in C_{2} ⟶ r_{1} \circ r_{2} = i \in C_{0} \\ r_{1} \in C_{1}, f_{2} \in C_{2} ⟶ r_{1} \circ f_{2} = f \in C_{0} \\ f_{3} \in C_{1}, r_{2} \in C_{2} ⟶ f_{3} \circ r_{2} = f_{2} \in C_{2} \\ f_{3} \in C_{1}, f_{2} \in C_{2} ⟶ f_{3} \circ f_{2} = r_{2} \in C_{2} \end{matrix}

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.

🔗

Observation 15.4.4.

🔗

This last example changes our course of action. If we had gotten a factor group from

{C_{0}, C_{1}, C_{2}},

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^{'} = (a * h_{1}) * (b * h_{2}) = (a * b) * (h_{1} * h_{2})

🔗

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^{'}

(a * b) * h

for some

h

H .

All that we really needed in the “abelian step” was that

h_{1} * b = b * (something in H) = b * h_{3} .

Then, since

H

is closed under

G

’s operation,

h_{3} * h_{2}

is an element of

H .

The consequence of this observation is that we define a certain kind of subgroup that guarantees that the inducted operation is well-defined.

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Definition 15.4.5. Normal Subgroup.

🔗

G

is a group,

H \leq G,

then

H

is a normal subgroup of

G,

denoted

H ◃ G,

if and only if every left coset of

H

is a right coset of

H;

i. e.

a * H = H * a \forall a \in G

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Theorem 15.4.6.

🔗

H \leq G,

then the operation induced on left cosets of

H

by the operation of

G

is well defined if and only if any one of the following conditions is true:

$H$ is a normal subgroup of $G .$
If $h \in H,$ $a \in G,$ then there exists $h^{'} \in H$ such that $h * a = a * h^{'} .$
If $h \in H,$ $a \in G,$ then $a^{- 1} * h * a \in H .$

Proof.

We leave the proof of this theorem to the reader.

🔗

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

🔗

Corollary 15.4.7.

🔗

H \leq G,

then the operation induced on left cosets of

H

by the operation of

G

is well defined if either of the following two conditions is true.

$G$ is abelian.
$| H | = \frac{| G |}{2} .$

Example 15.4.8. A non-normal subgroup.

The right cosets of

⟨ f_{1} ⟩ \leq S_{3}

are

{i, f_{1}},

{r_{1} f_{2}},

and

{r_{2}, f_{3}} .

These are not the same as the left cosets of

⟨ f_{1} ⟩ .

In addition,

f_{2}^{- 1} f_{1} f_{2} = f_{2} f_{1} f_{2} = f_{3} \notin ⟨ f_{1} ⟩ .

Thus,

⟨ f_{1} ⟩

is not normal.

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The improper subgroups

{e}

and

G

of any group

G

are normal subgroups.

G / {e}

is isomorphic to

G .

All other normal subgroups of a group, if they exist, are called proper normal subgroups.

Example 15.4.9.

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

Z_{2} .

Example 15.4.10. Subgroups 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,

σ_{1} σ_{2} \dots σ_{k},

where the length of

σ_{i}

ℓ_{i},

then for any

g \in S_{n},

g^{- 1} \circ f \circ g,

which is the conjugate of

f

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{aligned} 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{aligned}

Notice that the condition for normality of a subgroup

H

G

is that the conjugate of any element of

H

by an element of

G

must be remain in

H .

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.

Example 15.4.11.

Let

G

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

G = {(\begin{array}{cc} a & b \\ c & d \end{array}) ∣ a, b, c, d \in R, a d - b c \neq 0}

We saw in Chapter 11 that

G

is a group with matrix multiplication.

This group has many subgroups, but consider just two:

H_{1} = {(\begin{array}{cc} a & 0 \\ 0 & a \end{array}) | a \neq 0}

and

H_{2} = {(\begin{array}{cc} a & 0 \\ 0 & d \end{array}) | a d \neq 0} .

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 .

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Subsection 15.4.2 Homomorphisms

🔗

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

θ (x * y) = θ (x) ⋄ θ (y)

🔗

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.

🔗

Definition 15.4.12. Homomorphism.

🔗

Let

[G; *]

and

[G^{'}; ⋄]

be groups.

θ : G \to G^{'}

is a homomorphism if

θ (x * y) = θ (x) ⋄ θ (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.

Example 15.4.13. Decreasing modularity.

Define

α : Z_{6} \to Z_{3}

α (n) = n mod 3 .

Therefore,

α (0) = 0,

α (1) = 1,

α (2) = 2,

α (3) = 1 + 1 + 1 = 0,

α (4) = 1,

and

α (5) = 2 .

n, m \in Z_{6} .

We could actually show that

α

is a homomorphism by checking all

6^{2} = 36

different cases for the formula

\begin{matrix} (15.4.1) & α (n +_{6} m) = α (n) +_{3} α (m) \end{matrix}

but we will use a line of reasoning that generalizes. We have already encountered Sun Tzu’s Theorem, which implies that the function

β : Z_{6} \to Z_{3} \times Z_{2}

defined by

β (n) = (n mod 3, n mod 2) .

We need only observe that equating the first coordinates of both sides of the equation

\begin{matrix} (15.4.2) & β (n +_{6} m) = β (n) + β (m) \end{matrix}

gives us precisely the homomorphism property.

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Theorem 15.4.14. Group Homomorphism Properties.

🔗

θ : G \to G^{'}

is a homomorphism, then:

$θ (e) = θ (the identity of G) = the identity of G^{'} = e^{'} .$
$θ (a^{- 1}) = θ (a)^{- 1}$ for all $a \in G .$
If $H \leq G,$ then $θ (H) = {θ (h) | h \in H} \leq G^{'} .$

Proof.

Let $a$ be any element of $G .$ Then $θ (a) \in G^{'} .$

$\begin{aligned} θ (a) ⋄ e^{'} & = θ (a) by the definition of e^{'} \\ = θ (a * e) by the definition of e \\ = θ (a) ⋄ θ (e) by the fact that θ is a homomorphism \end{aligned}$

By cancellation, $e^{'} = θ (e) .$
Again, let $a \in G .$ $e^{'} = θ (e) = θ (a * a^{- 1}) = θ (a) ⋄ θ (a^{- 1}) .$ Hence, by the uniqueness of inverses, $θ (a)^{- 1} = θ (a^{- 1}) .$
Let $b_{1}, b_{2} \in θ (H) .$ Then there exists $a_{1}, a_{2} \in H$ such that $θ (a_{1}) = b_{1},$ $θ (a_{2}) = b_{2} .$ Recall that a compact necessary and sufficient condition for $H \leq G$ is that $x * y^{- 1} \in H$ for all $x, y \in H .$ Now we apply the same condition in $G^{'} :$

$\begin{aligned} b_{1} ⋄ b_{2}^{- 1} & = θ (a_{1}) ⋄ θ (a_{2})^{- 1} \\ = θ (a_{1}) ⋄ θ (a_{2}^{- 1}) \\ = θ (a_{1} * a_{2}^{- 1}) \in θ (H) \end{aligned}$

since $a_{1} * a_{2}^{- 1} \in H,$ and so we can conclude that $θ (H) \leq G^{'} .$

🔗

Corollary 15.4.15.

🔗

Since a homomorphism need not be a surjection and part (c) of Theorem 15.4.14 is true for the case of

H = G,

the range of

θ,

θ (G),

is a subgroup of

G^{'}

Example 15.4.16.

If we define

π : Z \to Z / 4 Z

π (n) = n + 4 Z,

then

π

is a homomorphism. The image of the subgroup

4 Z

is the single coset

0 + 4 Z,

the identity of the factor group. Homomorphisms of this type are called natural homomorphisms. The following theorems will verify that

π

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.

🔗

Theorem 15.4.17.

🔗

H ◃ G,

then the function

π : G \to G / H

defined by

π (a) = a H

is a homomorphism.

Proof.

We leave the proof of this theorem to the reader.

🔗

Definition 15.4.18. Natural Homomorphism.

🔗

H ◃ G,

then the function

π : G \to G / H

defined by

π (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.

🔗

Definition 15.4.19. Kernel of a homomorphism.

🔗

Let

θ : G \to G^{'}

be a homomorphism, and let

e

and

e^{'}

be the identities of

G

and

G^{'},

respectively. The kernel of

θ

is the set

\ker θ = {a \in G ∣ θ (a) = e^{'}}

🔗

Theorem 15.4.20.

🔗

Let

θ : G \to G^{'}

be a homomorphism from

G

into

G^{'}

. The kernel of

θ

is a normal subgroup of

G .

Proof.

Let

K = ker θ .

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

θ (a * b^{- 1}) = θ (a) * θ (b)^{- 1} = e^{'} * e^{' - 1} = e^{'} .

To prove normality, we let

g

be any element of

G

and

k \in K .

We compute

θ (g * k * g^{- 1})

to verify that

g * k * g^{- 1} \in K .

\begin{aligned} θ (g * k * g^{- 1}) & = θ (g) * θ (k) * θ (g^{- 1}) \\ = θ (g) * θ (k) * θ (g)^{- 1} \\ = θ (g) * e^{'} * θ (g)^{- 1} \\ = θ (g) * θ (g)^{- 1} \\ = e^{'} \end{aligned}

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

🔗

Theorem 15.4.21. Fundamental Theorem of Group Homomorphisms.

🔗

Let

θ : G \to G^{'}

be a homomorphism. Then

θ (G)

is isomorphic to

G / \ker θ .

Example 15.4.22.

Define

θ : Z \to Z_{10}

θ (n) = n mod 10 .

The three previous theorems imply the following:

$π : Z \to Z / 10 Z$ defined by $π (n) = n + 10 Z$ is a homomorphism.
${n \in Z | θ (n) = 0} = {10 n ∣ n \in Z} = 10 Z ◃ Z .$
$Z / 10 Z$ is isomorphic to $Z_{10} .$

Example 15.4.23.

Let

G

be the same group of two by two invertible real matrices as in Example 15.4.11. Define

Φ : G \to G

Φ (A) = \frac{A}{\sqrt{| det A |}} .

We will let the reader verify that

Φ

is a homomorphism. The theorems above imply the following.

$\ker Φ = {A \in G | Φ (A) = I} = {(\begin{array}{cc} a & 0 \\ 0 & a \end{array}) ∣ a \in R, a \neq 0} ◃ G .$ This verifies our statement in Example 15.4.11. As in that example, let $\ker Φ = H_{1} .$
$G / H_{1}$ is isomorphic to ${A \in G ∣ det A = 1} .$
$π : G \to G / H_{1}$ defined, naturally, by $π (A) = A H_{1}$ 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.

Example 15.4.24.

Consider

Φ : Z_{2}^{2} \to Z_{2}^{3}

defined by

Φ (a, b) = (a, b, a +_{2} b) .

(a_{1}, b_{1}), (a_{2}, b_{2}) \in Z_{2}^{2},

\begin{aligned} Φ ((a_{1}, b_{1}) + (a_{2}, b_{2})) & = Φ (a_{1} +_{2} a_{2}, b_{1} +_{2} b_{2}) \\ = (a_{1} +_{2} a_{2}, b_{1} +_{2} b_{2}, a_{1} +_{2} a_{2} +_{2} b_{1} +_{2} b_{2}) \\ = (a_{1}, b_{1}, a_{1} +_{2} b_{1}) + (a_{2}, b_{2}, a_{2} +_{2} b_{2}) \\ = Φ (a_{1}, b_{1}) + Φ (a_{2}, b_{2}) \end{aligned}

Since

Φ (a, b) = (0, 0, 0)

implies that

a = 0

and

b = 0,

the kernel of

Φ

{(0, 0)} .

By previous theorems,

Φ (Z_{2}^{2}) = {(0, 0, 0), (1, 0, 1), (0, 1, 1), (1, 1, 0)}

is isomorphic to

Z_{2}^{2} .

🔗

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

Z_{2}

), then

Φ : Z_{2}^{m} \to Z_{2}^{n}

defined by

Φ (a_{1}, a_{2}, . . ., a_{m}) = (a_{1}, a_{2}, . . ., a_{m}) 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

Z_{2} .

a = (a_{1}, a_{2}, . . ., a_{m})

and

b = (b_{1}, b_{2}, . . ., b_{m}),

(a + b) A = a A + b A

is true by basic matrix laws. Therefore,

Φ (a + b) = Φ (a) + Φ (b) .

🔗

Exercises 15.4.3 Exercises

🔗

1.

🔗

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

$θ_{1} : R^{*} \to R^{+}$ defined by $θ_{1} (a) = | a | .$
$θ_{2} : Z_{5} \to Z_{2}$ where $θ_{2} (n) = {\begin{array}{cc} 0 & if n is even \\ 1 & if n is odd \end{array} .$
$θ_{3} : R \times R \to R,$ where $θ_{3} (a, b) = a + b .$
$θ_{4} : S_{4} \to S_{4}$ defined by $θ_{4} (f) = f \circ f = f^{2} .$

Answer.

Yes, the kernel is ${1, - 1}$
No, since $θ_{2} (2 +_{5} 4) = θ_{2} (1) = 1,$ but $θ_{2} (2) +_{2} θ_{2} (4) = 0 +_{2} 0 = 0$

A follow-up might be to ask what happens if 5 is replaced with some other positive integer in this part.
Yes, the kernel is ${(a, - a) | a \in R}$
No. A counterexample, among many, would be to consider the two transpositions $t_{1} = (1, 3)$ and $t_{2} = (1, 2) .$ Compare $θ_{4} (t_{1} \circ t_{2})$ and $θ_{4} (t_{1}) \circ θ_{4} (t_{2}) .$

🔗

2.

🔗

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

$α_{1} : M_{2 \times 2} (R) \to R,$ defined by $α_{1} (A) = A_{11} A_{22} + A_{12} A_{21} .$
$α_{2} : {(R^{*})}^{2} \to R^{*}$ defined by $α_{2} (a, b) = a b .$
$α_{3} : {A \in M_{2 \times 2} (R) | det A \neq 0} \to R^{*},$ where $α_{3} (A) = det A .$
$α_{4} : S_{4} \to S_{4}$ defined by $α_{4} (f) = f^{- 1} .$

🔗

3.

🔗

Show that

D_{4}

has one proper normal subgroup, but that

⟨ (1, 4) (2, 3) ⟩

is not normal.

Answer.

⟨ r ⟩ = {i, r, r^{2}, r^{3}}

is a normal subgroup of

D_{4} .

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 ⟨ r ⟩

for all

a \in D_{4}

and

h \in ⟨ r ⟩ .

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.

⟨ f_{1} ⟩

is not a normal subgroup of

D_{4} .

⟨ f_{1} ⟩ = {i, f_{1}}

and if we choose

a = r

and

h = f_{1}

then

a^{- 1} h a = r^{3} f_{1} r = f_{2} \notin ⟨ f_{1} ⟩

🔗

4.

🔗

Prove that the function

Φ

in Example 15.4.23 is a homomorphism.

🔗

5.

🔗

Define the two functions

α : Z_{2}^{3} \to Z_{2}^{4}

and

β : Z_{2}^{4} \to Z_{2}

α (a_{1}, a_{2}, a_{3}) = (a_{1}, a_{2}, a_{3}, a_{1} +_{2} a_{2} +_{2} a_{3}),

and

β (b_{1}, b_{2}, b_{3}, b_{4}) = b_{1} + b_{2} + b_{3} + b_{4}

Describe the function

β \circ α .

Is it a homomorphism?

Answer.

(β \circ α) (a_{1}, a_{2}, a_{3}) = 0

and so

β \circ α

is the trivial homomorphism, but a homomorphism nevertheless.

🔗

6.

🔗

Express

Φ

in Example 15.4.23 in matrix form.

🔗

7.

🔗

Prove that if

G

is an abelian group, then

q (x) = x^{2}

defines a homomorphism from

G

into

G .

q

ever an isomorphism?

Answer.

Let

x, y \in G .

\begin{aligned} q (x * y) & = (x * y)^{2} \\ = x * y * x * y \\ = x * x * y * y since G is  abelian \\ = x^{2} * y^{2} \\ = q (x) * q (y) \end{aligned}

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{aligned} x \in Ker (q) & \Leftrightarrow q (x) = e \\ \Leftrightarrow x * x = e \\ \Leftrightarrow x^{- 1} = x \end{aligned}

🔗

8.

🔗

Prove that if

θ : G \to G^{'}

is a homomorphism, and

H ◃ G,

then

θ (H) ◃ θ (G) .

Is it also true that

θ (H) ◃ G^{'} ?

🔗

9.

🔗

Prove that if

θ : G \to G^{'}

is a homomorphism, and

H^{'} \leq θ (G),

then

θ^{- 1} (H^{'}) = {a \in G | θ (a) \in H^{'}} \leq G .

Answer.

Proof: Recall that the inverse image of

H^{'}

under

θ

θ^{- 1} (H^{'}) = {g \in G | θ (g) \in H^{'}} .

Closure: Let

g_{1}, g_{2} \in θ^{- 1} (H^{'}),

then

θ (g_{1}), θ (g_{2}) \in H^{'} .

Since

H^{'}

is a subgroup of

G^{'},

θ (g_{1}) ⋄ θ (g_{2}) = θ (g_{1} * g_{2}) \in H^{'} \Rightarrow g_{1} * g_{2} \in θ^{- 1} (H^{'})

Identity: By Theorem 15.4.14(a),

e \in θ^{- 1} (H^{'}) .

Inverse: Let

a \in θ^{- 1} (H^{'})

. Then

θ (a) \in H^{'}

and by Theorem 15.4.14(b),

θ (a)^{- 1} = θ (a^{- 1}) \in H^{'}

and so

a^{- 1} \in θ^{- 1} (H^{'}) .

🔗

10.

🔗

Following up on Example 15.4.10, prove that

A_{5}

is a simple group; i. e., it has no proper normal subgroups.

Make a list of the different cycle structures that occur in $A_{5}$ and how many elements have those structures.
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, $C,$ $f, g \in C$ if and only if $\exists ϕ \in A_{5}$ such that $ϕ^{- 1} \circ f \circ ϕ = g .$
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.

You have attempted of activities on this page.

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Section 15.4 Normal Subgroups and Group Homomorphisms

Subsection 15.4.1 Normal Subgroups

Example 15.4.1. Cosets of A3.

Example 15.4.3. Cosets of another subgroup of S3.

Observation 15.4.4.

Definition 15.4.5. Normal Subgroup.

Theorem 15.4.6.

Proof.

Corollary 15.4.7.

Example 15.4.8. A non-normal subgroup.

Example 15.4.9.

Example 15.4.10. Subgroups of A5.

Example 15.4.11.

Subsection 15.4.2 Homomorphisms

Definition 15.4.12. Homomorphism.

Example 15.4.13. Decreasing modularity.

Theorem 15.4.14. Group Homomorphism Properties.

Proof.

Corollary 15.4.15.

Example 15.4.16.

Theorem 15.4.17.

Proof.

Definition 15.4.18. Natural Homomorphism.

Definition 15.4.19. Kernel of a homomorphism.

Theorem 15.4.20.

Proof.

Theorem 15.4.21. Fundamental Theorem of Group Homomorphisms.

Example 15.4.22.

Example 15.4.23.

Example 15.4.24.

Exercises 15.4.3 Exercises

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Example 15.4.1. Cosets of $A_{3}$ .

Example 15.4.3. Cosets of another subgroup of $S_{3}$ .

Example 15.4.10. Subgroups of $A_{5}$ .