Theorem11.10First Isomorphism Theorem

If $\psi : G \rightarrow H$ is a group homomorphism with $K =\ker \psi\text{,}$ then $K$ is normal in $G\text{.}$ Let $\phi: G \rightarrow G/K$ be the canonical homomorphism. Then there exists a unique isomorphism $\eta: G/K \rightarrow \psi(G)$ such that $\psi = \eta \phi\text{.}$

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,

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

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, $g_1K =g_2K\text{.}$

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

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

Theorem11.12Second Isomorphism Theorem

Let $H$ be a subgroup of a group $G$ (not necessarily normal in $G$) and $N$ a normal subgroup of $G\text{.}$ Then $HN$ is a subgroup of $G\text{,}$ $H \cap N$ is a normal subgroup of $H\text{,}$ and

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

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

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

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

Since

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

Theorem11.13Correspondence Theorem

Let $N$ be a normal subgroup of a group $G\text{.}$ Then $H \mapsto H/N$ is a one-to-one correspondence between the set of subgroups $H$ containing $N$ and the set of subgroups of $G/N\text{.}$ Furthermore, the normal subgroups of $G$ containing $N$ correspond to normal subgroups of $G/N\text{.}$

Proof

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

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

Theorem11.14Third Isomorphism Theorem

Let $G$ be a group and $N$ and $H$ be normal subgroups of $G$ with $N \subset H\text{.}$ Then

Example11.15

By the Third Isomorphism Theorem,

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