
## Section11.2The 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. \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 ). \end{equation*}

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

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). \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, but

\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 ). \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 $\psi = \eta \phi\text{.}$

###### Example11.11

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). \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 ${\mathbb Z}_n\text{.}$

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} ). \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'). \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. \end{equation*}

Since

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

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

Let $H$ be a subgroup of $G$ containing $N\text{.}$ Since $N$ is normal in $H\text{,}$ $H/N$ makes sense. 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 11.13, we have also proved the following theorem.

###### Example11.15

By the Third Isomorphism Theorem,

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

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