§10 Isomorphism Theorems

§10.1 The First Isomorphism Theorem

§10.1.1 The Quotient Map as a Group Homomorphism

Proposition 10.1 Let $H\subset G$ be a normal subgroup. The map

\begin{aligned} q:G&\to G/H\\ g&\mapsto Hg \end{aligned}

(1) is a group homomorphism.

(2) is surjective.

(3) has kernel $q$ .

Proof:

(1)

\begin{aligned} q(g_{1}g_{2}) &=Hg_{1}g_{2}\\ &=Hg_{1}Hg_{2}\\ &=q(g_{1})q(g_{2}). \end{aligned}

(2) For every $Hg\in G/H$ ,

Hg=q(g).

(3)

q(g)=1_{G/H} \Longleftrightarrow \begin{aligned} q(g)&=H1_{G}\\ &=H. \end{aligned}

But

Hg=H1_{G}

if and only if $g$ and $1_G$ lie in the same orbit,

if and only if

g=h1_G

for some $h\in H$ ,

if and only if

g\in H.

Therefore,

q(g)=1_{G/H} \Longleftrightarrow g\in H.

~\tag*{$\square$}

§10.1.2 Visualization

What does this look like?

Think of $G$ as some set:

The subgroup $H\subset G$ partitions $G$ into orbits:

Then $G/H$ collapses each of these orbits to a single point:

§10.1.3 Injectivity and the Kernel of a Group Homomorphism

By the way:

Proposition 10.2 Let

G\xrightarrow{\phi}G^{\prime}

be a group homomorphism. Then $\phi$ is injective if and only if

\ker(\phi)=\{1_G\}.

Proof: If $\phi$ is injective, then there exists at most one $g$ such that

\phi(g)=1_{G^{\prime}}.

Since every group homomorphism sends $1_G$ to $1_{G^{\prime}}$ , we have

g\in 1_G.

Conversely, suppose

\ker(\phi)=\{1_G\}.

Then

\begin{aligned} \phi(g_1)=\phi(g_2) &\Rightarrow \phi(g_1)\phi(g_2)^{-1}=1_{G^{\prime}}\\ &\Rightarrow \phi(g_1g_2^{-1})=1_{G^{\prime}}\\ &\Rightarrow g_1g_2^{-1}\in\ker(\phi)\\ &\Rightarrow g_1g_2^{-1}=1_G\\ &\Rightarrow g_1=g_2. \end{aligned}

~\tag*{$\square$}

Thus, given any normal subgroup $H$ , the quotient homomorphism $q$ realizes $H$ as the kernel of some group homomorphism.

§10.1.4 Kernels Are Normal Subgroups

Question: Is the kernel of every group homomorphism a normal subgroup?

Proposition 10.3 Let

\phi:G\to G^{\prime}

be a group homomorphism. Then

\ker(\phi)

is a normal subgroup.

Proof: We need to show that for every

h\in\ker(\phi)

and every

g\in G,

we have

ghg^{-1}\in\ker(\phi).

Indeed,

\begin{aligned} \phi(ghg^{-1}) &=\phi(g)\phi(h)\phi(g^{-1})\\ &=\phi(g)1_{G^{\prime}}\phi(g^{-1})\\ &=\phi(g)\phi(g^{-1})\\ &=\phi(gg^{-1})\\ &=\phi(1_G)\\ &=1_{G^{\prime}}. \end{aligned}

Hence

ghg^{-1}\in\ker(\phi).

~\tag*{$\square$}

Here we have only proved

g\ker(\phi)g^{-1} \subset \ker(\phi), \qquad \forall g,

in order to conclude that $\ker(\phi)$ is normal.

But how do we prove

g\ker(\phi)g^{-1} = \ker(\phi)?

§10.1.5 Inclusion under Conjugation Implies Normality

Proposition 10.4 Let $H\subset G$ be a subgroup. “ $\forall g\in G$ , $gHg^{-1}\subset H$ ” implies “ $\forall g\in G$ , $gHg^{-1}=H$ ”. Proof: We need to show that $\forall g\in G$ , $H\subset gHg^{-1}$ .

So fix $h\in H$ . Let $g^{\prime}=g^{-1}$ . By hypothesis,

g^{\prime}H(g^{\prime})^{-1}\subset H,

g^{\prime}h(g^{\prime})^{-1}=h^{\prime}

for some $h^{\prime}\in H$ . Then

h=gh^{\prime}h^{-1}

Since

\begin{aligned} gh^{\prime}g^{-1}&=g(g^{\prime}h(g^{\prime})^{-1})g^{-1}\\ &=gg^{-1}hgg^{-1}\\ &=h \end{aligned}

This shows $h\in gHg^{-1}$ .

~\tag*{$\square$}

§10.1.6 Intersections of Normal Subgroups

Corollary 10.5 If $H_1$ and $H_2$ are normal subgroups of $G$ , then

H_1\cap H_2

is also a normal subgroup.

Proof: Let

h\in H_1\cap H_2.

Then for every $g\in G$ ,

$ghg^{-1}\in H_1$ because $H_1$ is normal;
$ghg^{-1}\in H_2$ because $H_2$ is normal.

Therefore,

ghg^{-1} \in H_1\cap H_2.

Hence, for every $g\in G$ ,

gH_1\cap H_2g^{-1} \subset H_1\cap H_2.

~\tag*{$\square$}

Notice also:

Proposition 10.6 If $H_1$ and $H_2$ are merely subgroups of $G$ , then

H_1\cap H_2

is also a subgroup of $G$ .

Proof: Since $H_1$ and $H_2$ are both subgroups,

1_G\in H_1

and

1_G\in H_2.

Therefore,

1_G\in H_1\cap H_2.

If $g$ and $g^{\prime}$ lie in $H_1\cap H_2$ , then

$gg^{\prime}\in H_1$ because $H_1$ is a subgroup;
$gg^{\prime}\in H_2$ because $H_2$ is a subgroup.

Therefore,

gg^{\prime}\in H_1\cap H_2.

The argument for inverses is similar.

~\tag*{$\square$}

§10.1.7 Constructing the Smallest Normal Subgroup Containing a Set

Suppose now that you have a group $G$ and an arbitrary set

I

whose elements lie in $G$ .

( $I$ need not be a subgroup; it is merely some arbitrary collection of elements of $G$ .)

Question: Can you find a normal subgroup of $G$ containing $I$ ?

Well, $G$ itself is a normal subgroup of $G$ , and it certainly contains $I$ .

Question: Can we find a smaller subgroup?

Yes. Consider the intersection

\bigcap H.

The set

{H\subset G\mid H~\text{is normal and}~H~\text{contains}~I}

is nonempty because $G$ itself belongs to it.

Therefore, by taking this intersection, we obtain a normal subgroup of $G$ by the previous corollary.

This is a very useful construction.

§10.1.8 The First Isomorphism Theorem

Proposition 10.7 Let $\phi:G\to G^{\prime}$ be a surjective group homomorphism. Then $\exists$ an isomorphism $G/\ker(\phi)\xrightarrow{\cong} G^{\prime}$ .

Proof: Given a surjective group homomorphism

\phi:G\to G^{\prime},

Let $H=\ker(\phi)$ . Note that if $g_{2}\in Hg_{1}$ ,

\phi(g_{2})=g_{1}

Because

\begin{aligned} \phi(g_{2})&=\phi(hg_{1}), \quad\text{for some}~h\in H\\ &=\phi(h)\phi(g_{1})\\ &=1_{G^{\prime}}\phi(g_{1})\\ &=\phi(g_{1}) \end{aligned}

So we have a well-defined map

\begin{aligned} \psi: G/H&\to G^{\prime}\\ Hg&\mapsto \phi(g) \end{aligned}

(We showed if $Hg_{1}=Hg_{2}$ , then $\phi(g_{1})=\phi(g_{2})$ .) This is a homomorphism, since

\begin{align*} \psi(Hg_{1}Hg_{2})&=\psi(Hg_{1}g_{2})\tag*{in $G/H$}\\ &=\phi(g_{1}g_{2})\tag*{Definition of $\psi$}\\ &=\phi(g_{1})\phi(g_{2})\tag*{$\phi$ is a homomorphism}\\ &=\psi(Hg_{1})\psi(Hg_{2})\tag*{Definition of $\psi$} \end{align*}

It is an injection, since

\begin{aligned} \psi(Hg_{1})=1_{G^{\prime}} & \Leftrightarrow \psi(g_{1})=1_{G^{\prime}}\\ &\Leftrightarrow g_{1}\in H\\ &\Leftrightarrow Hg_{1}=H1_{G},~\text{the unit of}~G/H \end{aligned}

It is a surjection since $\phi$ is a surjection:

\forall g^{\prime}\in G^{\prime}, \exists~\text{some}~g\in G, \text{s.t.}~\phi(g)=g^{\prime}, \text{so}~\psi(Hg)=g.

~\tag*{$\square$}

Corollary 10.8 (First Isomorphism Theorem)

Let $\phi:G\to G^{\prime}$ be any group homomorphism. Then $\exists$ group isomorphism

G/\ker(\phi)\cong \mathrm{im}(\phi).

Proof: $\mathrm{im}(\phi)\subset G^{\prime}$ is a subgroup. By the definition of image, the homomorphism $\phi: G\to G^{\prime}$ factors as follows:

Here, $j$ is the inclusion of $\mathrm{im}(\phi)$ into $G$ . (It's a injective group homomorphism.) $\overline{\phi}$ is the “same” function as $\phi$ , but has a different target/codomain. So we see $\phi=j\circ \overline{\phi}$ .

Also, by definition of image, $\overline{\phi}$ is a surjection. Hence, the proposition says

G/\ker(\overline{\phi})\cong \mathrm{im}(\phi)

But $\ker(\overline{\phi})=\ker(\phi)$ , since $j(1_{\mathrm{im}(\phi)})=1_{G^{\prime}}$ .

~\tag*{$\square$}

§10.1.9 Application of the First Isomorphism Theorem: Index

Proposition 10.9 Let

O_{n}(\mathbb{R}):=\{n\times n~\text{real matrices}~A~\text{such that}~A^{\mathrm{T}}A=I\}

and

SO_{n}(\mathbb{R}):=\{A\in O_{n}(\mathbb{R})~\text{s.t.}~\det(A)=1\}.

Then

[O_{n}(\mathbb{R}):SO_{n}(\mathbb{R})]=2.

i.e. $SO_{n}(\mathbb{R})$ is an index $2$ subgroup of $O_{n}(\mathbb{R})$ .

Proof: Consider the homomorphism

\begin{aligned} \det:O_{n}(\mathbb{R})&\to \mathbb{R}^{\times}\\ A&\mapsto \det(A) \end{aligned}

Since the unit of $\mathbb{R}^{\times}$ is $1\in\mathbb{R}^{\times}$ , the kernel of $\det$ is $SO_{n}(\mathbb{R})$ .

On the other hand,

\mathrm{im}(\det)=\{+1,-1\}\subset \mathbb{R}^{\times}.

Since

\begin{align*} [O_{n}(\mathbb{R}):SO_{n}(\mathbb{R})]&=|O_{n}(\mathbb{R})/SO_{n}(\mathbb{R})|\tag*{by definition of index}\\ &=|O_{n}(\mathbb{R})/\ker(\det)|\\ &=|\mathrm{im}(\det)|\tag*{The First Isomorphism Theorem}\\ &=|\{+1,-1\}|\\ &=2 \end{align*}

~\tag*{$\square$}

§10.2 The Second Isomorphism Theorem

§10.2.1 The Second Isomorphism Theorem

Fix a group $G$ . Let $S\subset G$ be a subgroup and $N\triangleleft G$ be a normal subgroup.

Proposition 10.10 Let $SN$ be the set of all elements in $G$ of the form $sx$ where $s\in S$ and $x\in N$ . This is a subgroup of $G$ .

Proof: Given $s_{1}, s_{2}\in S$ and $x_{1}, x_{2}\in N$ , we have that

s_{1}x_{1}s_{2}x_{2}=s_{1}s_{2}s_{2}^{-1}x_{1}s_{2}x_{2}=s_{1}s_{2}x^{\prime} x_{2}

for some $x^{\prime}\in N$ (since $N$ is normal). And $s_{1}s_{2}\in S$ and $x^{\prime}x_{2}\in N$ since both are closed under multiplication. The identity is in $SN$ since $1\in S$ , $N$ and $1\cdot 1=1$ . Finally, $SN$ contains inverses because

x^{-1}s^{-1}=(s^{-1}x^{\prime}s)s^{-1}=s^{-1}x^{\prime}

where $x^{\prime}\in N$ is the element such that $x^{\prime}=sx^{-1}s^{-1}$ .

~\tag*{$\square$}

Proposition 10.11 $N$ is a normal subgroup of $SN$ .

Proof: We know that for every $g\in G$ and $x\in N$ ,

gxg^{-1}\in N.

Since

SN\subset G,

in particular, for every $g\in SN$ ,

gxg^{-1}\in N.

~\tag*{$\square$}

Proposition 10.12

S\cap N

is a normal subgroup of $S$ .

Proof: If

x\in S\cap N,

then for every $s\in S$ , we know

sxs^{-1}\in N

because $N$ is normal in $G$ .

On the other hand, since $S$ is closed under multiplication,

sxs^{-1}\in S

as well.

Therefore,

sxs^{-1} \in N\cap S.

~\tag*{$\square$}

Theorem 10.13 (Second Isomorphism Theorem)

There exists an isomorphism

S/(S\cap N) \cong SN/N.

Proof: Consider the composition of homomorphisms

S\to SN\to SN/N,

where the second map is the quotient map and the first is simply the inclusion map.

Notice that

S\subset SN.

This composition is surjective because for any $n\in N$ , the element

[sn]\in SN/N

is equivalent to

[s]\in SN/N.

Its kernel consists of precisely those elements of $S$ that lie in $N$ , namely

S\cap N.

Therefore, the result follows from the First Isomorphism Theorem.

~\tag*{$\square$}

Question: Does the coset $[s]$ in

S/(S\cap N)

define the coset $[sn]$ in

SN/N?

Does the choice of $n$ in $[sn]$ matter?

This suggests another proof of the Second Isomorphism Theorem.

Proof: Given

[sn]\in SN/N,

consider

[s]\in S/(S\cap N).

We claim that the assignment

\phi:[sn]\mapsto[s]

is well defined.

sn=s^{\prime}n^{\prime}x

with $x\in N$ , then

# s s^{\prime}(n^{\prime}xn^{-1}).

We must show that

n^{\prime}xn^{-1} \in S\cap N.

By multiplying both sides of the equation on the left by $s^{\prime-1}$ , we see that it must lie in $S$ .

We also know that it lies in $N$ because $n^{\prime}$ , $x$ , and $n^{-1}$ all lie in $N$ , and $N$ is closed under multiplication.

Now we prove that this is a group homomorphism:

\begin{aligned} \phi([s_1n_1][s_2n_2]) &= \phi([s_1n_1s_2n_2])\ &= \phi([s_1s_2(s_2^{-1}n_1s_2n_2)])\ &= \phi([s_1s_2(n^{\prime}s_2)])\ &= [s_1s_2]\ &= [s_1][s_2]\ &= \phi([s_1n_1])\phi([s_2n_2]). \end{aligned}

To prove injectivity, we must show that the kernel is trivial.

\phi([sn])=[x]

for

x\in S\cap N,

then $[sn]$ has a representative of the form

xn^{\prime}.

But

x\in X\cap N, \qquad n^{\prime}\in N

implies

xn^{\prime}\in N,

since $N$ is closed under multiplication.

Therefore,

# [sn] # [sn^{\prime}] 1 \in SN/N.

To prove surjectivity, observe that for every $s\in S$ ,

s=s1_G\in SN.

Thus

# \phi([s1_G]) \phi(s).

~\tag*{$\square$}

§10.2.2 Application of the Second Isomorphism Theorem

Example 10.1 Let

G=S_4,

the symmetric group on $4$ elements.

Let $S$ be the subgroup generated by the permutation

(12),

and let

N=A_4

be the alternating group on $4$ elements.

We have an isomorphism

S\cong\mathbb{Z}_2.

Proof:

Determine $S$ , $N$ , and their intersection.

S=\langle(12)\rangle,

which has order $2$ .

N=A_4,

which has order $12$ .

For

S\cap N:

since $(12)$ is an even permutation only if it can be expressed as an even number of transpositions, but $(12)$ is itself a single transposition,

(12)\notin A_4.

Therefore,

# S\cap N {1_G}.

Compute $SN$ .

$SN$ consists of all elements that can be written as

sn,

where $s\in S$ and $n\in N$ .

Since

|S|=2, \qquad |N|=12, \qquad S\cap N={1_G},

the order of $SN$ is

# \frac{|S||N|}{|S\cap N|} # \frac{2\times12}{1} 24.

But

|G|=24,

SN=G.

Apply the Second Isomorphism Theorem:

S \cong S/{1_G} \cong S/(S\cap N) \cong (SN)/N \cong G/N \cong S_4/A_4 \cong \mathbb{Z}_2.

~\tag*{$\square$}

§10.3 The Third Isomorphism Theorem

The Third Isomorphism Theorem answers the following question.

Suppose I have a nested sequence of subgroups

K\subset N\subset G.

I can quotient out the whole subgroup $N$ at once and obtain the orbit set

G/N.

In doing so, $K$ is also quotiented out because

K\subset N.

Alternatively, I can quotient in stages: first take

G/K,

and then quotient out what remains of $N$ .

Do we obtain the same result?

The answer is yes. If both $K$ and $N$ are normal subgroups of $G$ , so that the relevant quotient groups are defined, then the final result is the same group.

§10.3.1 The Third Isomorphism Theorem

Proposition 10.14 Suppose there are subgroups

K\subset N\subset G.

There exists an injection

f:N/K\to G/K.

Proof: For any coset

nK

in $N/K$ , where $n\in N$ , define

f(nK)=nK\in G/K.

This simply means that we regard the coset $nK$ coming from $N/K$ as an element of $G/K$ .

Well-definedness: If

n_1K=n_2K

in $N/K$ , then

n_1^{-1}n_2\in K.

Since

K\subset G,

this also means

n_1K=n_2K

in $G/K$ .

Therefore, $f$ is well defined.

f(n_1K)=f(n_2K),

then

n_1K=n_2K

in $G/K$ , which means

n_1^{-1}n_2\in K.

Hence

n_1K=n_2K

in $N/K$ .

Therefore, $f$ is injective.

Thus there exists an injection

f:N/K\to G/K.

~\tag*{$\square$}

These are only maps between sets, not yet between groups. After all, we have not yet assumed that $K$ is normal in $G$ .

Proposition 10.15 Let

K\subset N\subset G

be subgroups, and suppose

K\triangleleft G.

Then

K\triangleleft N.

Proof: Since

K\triangleleft G,

for every $g\in G$ and $k\in K$ ,

gkg^{-1}\in K.

In particular, this holds for every $n\in N$ , since

N\subset G.

Thus, for every $n\in N$ and $k\in K$ ,

nkn^{-1}\in K.

Therefore $K$ is normal in $N$ , so

K\triangleleft N.

~\tag*{$\square$}

Now we can regard

G/K

and

N/K

as groups.

Proposition 10.16 Let

K\subset N\subset G

be subgroups.

The injection

f:N/K\to G/K

is a group homomorphism.

Proof: For any

n_1K,n_2K\in N/K,

their product in $N/K$ is

# (n_1K)\cdot(n_2K) (n_1n_2)K.

Applying $f$ gives

# f((n_1K)\cdot(n_2K)) # f((n_1n_2)K) (n_1n_2)K.

On the other hand,

# f(n_1K)\cdot f(n_2K) # (n_1K)\cdot(n_2K) (n_1n_2)K.

Since the two sides are equal, $f$ is a group homomorphism.

~\tag*{$\square$}

This shows that

N/K

is a subgroup of

G/K.

Proposition 10.17 Let

K\subset N\subset G

be subgroups.

There exists a bijection

\psi:G/N\to(G/K)/(N/K).

Proof: For any coset

gN

in $G/N$ , define

# \psi(gN) gK \in (G/K)/(N/K).

This means that $gK$ is regarded as a coset in $G/K$ with respect to the subgroup

N/K.

Well-definedness: If

g_1N=g_2N,

then

g_1^{-1}g_2\in N.

Therefore $g_1K$ and $g_2K$ represent the same coset in

(G/K)/(N/K).

Hence $\psi$ is well defined.

Injectivity: If

# \psi(g_1N) \psi(g_2N),

then $g_1K$ and $g_2K$ are the same double coset in

(G/K)/(N/K),

which means

g_1N=g_2N.

Therefore $\psi$ is injective.

Surjectivity: For every coset $gK$ in

(G/K)/(N/K),

there exists a corresponding

gN\in G/N

such that

\psi(gN)=gK.

Hence $\psi$ is surjective.

Therefore, $\psi$ is a bijection.

~\tag*{$\square$}

This is only a function between two sets.

To be explicit, on the right-hand side we use the action of

N/K

G/K,

since $N/K$ is a subgroup.

The quotient set

(G/K)/(N/K)

is the usual orbit space for this action.

Proposition 10.18 Let $G$ be a finite group and

K\subset N\subset G

be subgroups.

Then

# |G/N| |G/K|/|N/K|.

Proof: $|G/N|$ is the number of cosets of $N$ in $G$ .

$|G/K|$ is the number of cosets of $K$ in $G$ .

$|N/K|$ is the number of cosets of $K$ in $N$ .

Each coset of $N$ in $G$ corresponds to

|N/K|

cosets of $K$ in $G$ .

Therefore,

# |G/K| |G/N|\cdot|N/K|.

Rearranging,

# |G/N| \frac{|G/K|}{|N/K|}.

~\tag*{$\square$}

Theorem 10.19 (Third Isomorphism Theorem)

Let $G$ be a group.

Suppose

K\triangleleft G

and

N\triangleleft G,

with

K\subset N\subset G.

Then the quotient group

N/K

is a normal subgroup of

G/K,

and

(G/K)/(N/K) \cong G/N.

Proof:

$N/K$ is a normal subgroup of $G/K$ .

Since $N$ is a subgroup of $G$ ,

N/K

is a subgroup of

G/K.

For any coset

gK\in G/K

and any

nK\in N/K,

consider the conjugate:

# (gK)(nK)(gK)^{-1} # gKnKg^{-1}K gng^{-1}K.

Since

N\triangleleft G,

we have

gng^{-1}\in N.

Therefore,

gng^{-1}\in N/K.

Hence,

(gK)(nK)(gK)^{-1} \in N/K.

Thus

N/K\triangleleft G/K.

Define the natural homomorphism

\phi:G/K\to G/N

\phi(gK)=gN.

This is well defined.

Indeed, if

gK=g^{\prime}K,

then

g^{-1}g^{\prime}\in K\subset N.

Therefore,

g^{-1}g^{\prime}\in N,

gN=g^{\prime}N.

Hence

# \phi(gK) # gN # g^{\prime}N \phi(g^{\prime}K).

$\phi$ is a surjective group homomorphism.

For any

gK,hK\in G/K,

# \phi((gK)(hK)) # \phi(ghK) # ghN # (gN)(hN) \phi(gK)\phi(hN).

Therefore, $\phi$ is a homomorphism.

For any

gN\in G/N,

there exists

gK\in G/K

such that

\phi(gK)=gN.

Hence $\phi$ is surjective.

Determine the kernel of $\phi$ .

# \ker(\phi) {gK\in G/K\mid\phi(gK)=N}.

The equation

\phi(gK)=gN=N

means

g\in N.

Therefore,

# \ker(\phi) # {gK\mid g\in N} N/K.

Apply the First Isomorphism Theorem.

Since

\phi:G/K\to G/N

is a surjective homomorphism with kernel

N/K,

we have

(G/K)/\ker(\phi) \cong \mathrm{im}(\phi) ================= G/N.

Since

\ker(\phi)=N/K,

we obtain

(G/K)/(N/K) \cong G/N.

~\tag*{$\square$}

§10.3.2 Application of the Third Isomorphism Theorem

Example 10.2 Let

G=\mathbb{Z}, \qquad N=4\mathbb{Z}, \qquad K=12\mathbb{Z}.

Then

# N/K 4\mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}_{3}.

Also,

# G/K \mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}_{12}.

Therefore,

# (G/K)/(N/K) (\mathbb{Z}*{12})/(\mathbb{Z}*{3}) \cong \mathbb{Z}_{4}.

Meanwhile,

# G/N \mathbb{Z}/4\mathbb{Z} \cong \mathbb{Z}_{4}.

Hence,

(G/K)/(N/K) \cong G/N \cong \mathbb{Z}_{4}.

Contents

§10 Isomorphism Theorems

§10.1 The First Isomorphism Theorem

§10.1.1 The Quotient Map as a Group Homomorphism

§10.1.2 Visualization

§10.1.3 Injectivity and the Kernel of a Group Homomorphism

§10.1.4 Kernels Are Normal Subgroups

§10.1.5 Inclusion under Conjugation Implies Normality

§10.1.6 Intersections of Normal Subgroups

§10.1.7 Constructing the Smallest Normal Subgroup Containing a Set

§10.1.8 The First Isomorphism Theorem

§10.1.9 Application of the First Isomorphism Theorem: Index

§10.2 The Second Isomorphism Theorem

§10.2.1 The Second Isomorphism Theorem

§10.2.2 Application of the Second Isomorphism Theorem

§10.3 The Third Isomorphism Theorem

§10.3.1 The Third Isomorphism Theorem

§10.3.2 Application of the Third Isomorphism Theorem