## Section4.3The Isomorphism Theorems

Although it is not evident at first, factor groups correspond exactly to homomorphic images, and we can use factor groups to study homomorphisms. We already know that with every group homomorphism $\phi: G \rightarrow H$ we can associate a normal subgroup of $G\text{,}$ $\ker \phi\text{.}$ The converse is also true; that is, every normal subgroup of a group $G$ gives rise to homomorphism of groups.

Let $H$ be a normal subgroup of $G\text{.}$ Define the natural or canonical homomorphism

\begin{equation*} \phi : G \rightarrow G/H \end{equation*}

by

\begin{equation*} \phi(g) = gH\text{.} \end{equation*}

This is indeed a homomorphism, since

\begin{equation*} \phi( g_1 g_2 ) = g_1 g_2 H = g_1 H g_2 H = \phi( g_1) \phi( g_2 )\text{.} \end{equation*}

The kernel of this homomorphism is $H\text{.}$ The following theorems describe the relationships between group homomorphisms, normal subgroups, and factor groups.

###### Proof.

We already know that $K$ is normal in $G\text{.}$ Define $\eta: G/K \rightarrow \psi(G)$ by $\eta(gK) = \psi(g)\text{.}$ We first show that $\eta$ is a well-defined map. If $g_1 K =g_2 K\text{,}$ then for some $k \in K\text{,}$ $g_1 k=g_2\text{;}$ consequently,

\begin{equation*} \eta(g_1 K) = \psi(g_1) = \psi(g_1) \psi(k) = \psi(g_1k) = \psi(g_2) = \eta(g_2 K)\text{.} \end{equation*}

Thus, $\eta$ does not depend on the choice of coset representatives and the map $\eta: G/K \rightarrow \psi(G)$ is uniquely defined since $\psi = \eta \phi\text{.}$ We must also show that $\eta$ is a homomorphism. Indeed,

\begin{align*} \eta( g_1K g_2K ) & = \eta(g_1 g_2K)\\ & = \psi(g_1 g_2)\\ & = \psi(g_1) \psi(g_2)\\ & = \eta( g_1K) \eta( g_2K )\text{.} \end{align*}

Clearly, $\eta$ is onto $\psi( G)\text{.}$ To show that $\eta$ is one-to-one, suppose that $\eta(g_1 K) = \eta(g_2 K)\text{.}$ Then $\psi(g_1) = \psi(g_2)\text{.}$ This implies that $\psi( g_1^{-1} g_2 ) = e\text{,}$ or $g_1^{-1} g_2$ is in the kernel of $\psi\text{;}$ hence, $g_1^{-1} g_2K = K\text{;}$ that is, $g_1K =g_2K\text{.}$

Mathematicians often use diagrams called commutative diagrams to describe such theorems. The following diagram “commutes” since $\psi = \eta \phi\text{.}$

###### Example4.23.

Let $G$ be a cyclic group with generator $g\text{.}$ Define a map $\phi : {\mathbb Z} \rightarrow G$ by $n \mapsto g^n\text{.}$ This map is a surjective homomorphism since

\begin{equation*} \phi( m + n) = g^{m+n} = g^m g^n = \phi(m) \phi(n)\text{.} \end{equation*}

Clearly $\phi$ is onto. If $|g| = m\text{,}$ then $g^m = e\text{.}$ Hence, $\ker \phi = m {\mathbb Z}$ and ${\mathbb Z} / \ker \phi = {\mathbb Z} / m {\mathbb Z} \cong G\text{.}$ On the other hand, if the order of $g$ is infinite, then $\ker \phi = 0$ and $\phi$ is an isomorphism of $G$ and ${\mathbb Z}\text{.}$ Hence, two cyclic groups are isomorphic exactly when they have the same order. Up to isomorphism, the only cyclic groups are ${\mathbb Z}$ and ${\mathbb Z}_n\text{.}$

###### Proof.

We will first show that $HN = \{ hn : h \in H, n \in N \}$ is a subgroup of $G\text{.}$ Suppose that $h_1 n_1, h_2 n_2 \in HN\text{.}$ Since $N$ is normal, $(h_2)^{-1} n_1 h_2 \in N\text{.}$ So

\begin{equation*} (h_1 n_1)(h_2 n_2) = h_1 h_2 ( (h_2)^{-1} n_1 h_2 )n_2 \end{equation*}

is in $HN\text{.}$ The inverse of $hn \in HN$ is in $HN$ since

\begin{equation*} ( hn )^{-1} = n^{-1 } h^{-1} = h^{-1} (h n^{-1} h^{-1} )\text{.} \end{equation*}

Next, we prove that $H \cap N$ is normal in $H\text{.}$ Let $h \in H$ and $n \in H \cap N\text{.}$ Then $h^{-1} n h \in H$ since each element is in $H\text{.}$ Also, $h^{-1} n h \in N$ since $N$ is normal in $G\text{;}$ therefore, $h^{-1} n h \in H \cap N\text{.}$

Now define a map $\phi$ from $H$ to $HN / N$ by $h \mapsto h N\text{.}$ The map $\phi$ is onto, since any coset $h n N = h N$ is the image of $h$ in $H\text{.}$ We also know that $\phi$ is a homomorphism because

\begin{equation*} \phi( h h') = h h' N = h N h' N = \phi( h ) \phi( h')\text{.} \end{equation*}

By the First Isomorphism Theorem, the image of $\phi$ is isomorphic to $H / \ker \phi\text{;}$ that is,

\begin{equation*} HN/N = \phi(H) \cong H / \ker \phi\text{.} \end{equation*}

Since

\begin{equation*} \ker \phi = \{ h \in H : h \in N \} = H \cap N\text{,} \end{equation*}

$HN/N = \phi(H) \cong H / H \cap N\text{.}$

###### Proof.

Let $H$ be a subgroup of $G$ containing $N\text{.}$ Since $N$ is normal in $H\text{,}$ $H/N$ makes is a factor group. Let $aN$ and $bN$ be elements of $H/N\text{.}$ Then $(aN)( b^{-1} N )= ab^{-1}N \in H/N\text{;}$ hence, $H/N$ is a subgroup of $G/N\text{.}$

Let $S$ be a subgroup of $G/N\text{.}$ This subgroup is a set of cosets of $N\text{.}$ If $H= \{ g \in G : gN \in S \}\text{,}$ then for $h_1, h_2 \in H\text{,}$ we have that $(h_1 N)( h_2 N )= h_1 h_2 N \in S$ and $h_1^{-1} N \in S\text{.}$ Therefore, $H$ must be a subgroup of $G\text{.}$ Clearly, $H$ contains $N\text{.}$ Therefore, $S = H / N\text{.}$ Consequently, the map $H \mapsto H/N$ is onto.

Suppose that $H_1$ and $H_2$ are subgroups of $G$ containing $N$ such that $H_1/N = H_2/N\text{.}$ If $h_1 \in H_1\text{,}$ then $h_1 N \in H_1/N\text{.}$ Hence, $h_1 N = h_2 N \subset H_2$ for some $h_2$ in $H_2\text{.}$ However, since $N$ is contained in $H_2\text{,}$ we know that $h_1 \in H_2$ or $H_1 \subset H_2\text{.}$ Similarly, $H_2 \subset H_1\text{.}$ Since $H_1 = H_2\text{,}$ the map $H \mapsto H/N$ is one-to-one.

Suppose that $H$ is normal in $G$ and $N$ is a subgroup of $H\text{.}$ Then it is easy to verify that the map $G/N \rightarrow G/H$ defined by $gN \mapsto gH$ is a homomorphism. The kernel of this homomorphism is $H/N\text{,}$ which proves that $H/N$ is normal in $G/N\text{.}$

Conversely, suppose that $H/N$ is normal in $G/N\text{.}$ The homomorphism given by

\begin{equation*} G \rightarrow G/N \rightarrow \frac{G/N}{H/N} \end{equation*}

has kernel $H\text{.}$ Hence, $H$ must be normal in $G\text{.}$

Notice that in the course of the proof of Theorem 4.25 , we have also proved the following theorem.

###### Example4.27.

By the Third Isomorphism Theorem,

\begin{equation*} {\mathbb Z} / m {\mathbb Z} \cong ({\mathbb Z}/ mn {\mathbb Z})/ (m {\mathbb Z}/ mn {\mathbb Z})\text{.} \end{equation*}

Since $| {\mathbb Z} / mn {\mathbb Z} | = mn$ and $|{\mathbb Z} / m{\mathbb Z}| = m\text{,}$ we have $| m {\mathbb Z} / mn {\mathbb Z}| = n\text{.}$

###### 1.

If $\phi:G \to H$ is a group homomorphism, what can you say about $\ker \phi\text{?}$

###### 2.

If $N$ is a normal subgroup of $G\text{,}$ must there be a homomorphism that has $N$ as its kernel? If so, what is it?

###### 3.

What is the point of showing that $\eta(g_1Kg_2) = \eta(g_1K)\eta(g_2K)$ in the proof of Theorem 4.22 ? Also, why is the proof not done directly after showing this?

###### 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).

### ExercisesPractice Problems

###### 1.

Describe all the homomorphisms from $\Z_{24}$ to $\Z_{18}\text{.}$

Hint

The kernel of a homomorphism must be a subgroup of the domain and the image must be a subgroup of the codomain. Further, a generator must map to a generator.

###### 2.

Prove that $\Z/\langle 5\rangle \cong \Z_5\text{.}$

###### 3.

Let $\FR$ be the set of all functions $f:\R to \R$ under addition. Let $B = \{f \in \FR \st f(0) = 0\}\text{.}$ Prove that $\FR/B \cong \R\text{.}$

Hint

Define a surjective homomorphism $\psi:\FR \to \R$ and use the FHT.

###### 4.

How many homomorphic images are there of $D_4\text{?}$

### ExercisesCollected Homework

###### C1.

Let $G$ be an abelian group. Let $H = \{g^2 \st g \in G\}$ and $K = \{g \in G \st g^2 = e\}\text{.}$

1. Find $H$ and $K$ for the example $G = U(7)\text{.}$

2. Use the Fundamental Homomorphism Theorem the relate the groups $H$ and $K\text{.}$ That is, find a surjective homomorphism from $G$ to one of them, such that the other is the kernel. Justify your claims (including that your function really is a homomorphism). Then explain how one is isomorphic to a quotient group involving the other.