chapter 9 summary


A subgroup $H$ of a group $G$ is normal in $G$ if $gH = Hg$ for all $g \in G$. That is, $H$ is normal iff the left and right cosets are equal. We write: $H \unlhd G$.

If $N$ is a normal subgroup of a group $G,$ then the cosets of $N$ in $G$ form a group $G/N$ under the operation $(aN)(bN) = abN.$This group is called the factor or quotient group of $G$ and $N$.

A homomorphism is a map $\phi :G \rightarrow H$ such that $\phi (g_1, g_2)=\phi(g_1)\phi(g_2)$ for all $g_1,g_2 \in G$.

Let $\phi:G\rightarrow H$ be homomorphism. Define the kernel of $\phi$ to be $ker\phi = \{g\inG:\phi(g)=e\}$.
for finite groups $|G|$/ $|H|$


Theorem 9.1
Let $N\leq\ G$, then the following are equivalent.
1.) The subgroup $N$ is normal in in $G$.
2.) For all $g\in\ G$, $gNg^{-1}\subset\ N$.
3.) For all $g\in\ G$, $gNg^{-1} = N$.

Theorem 9.2
Let $N$ be normal to $G$. Then the cosets of $N$ form a group, denoted $G/N$ with the following operation:
$(aN)(bN) = abN$

Proposition 9.3: Let $\phi : G \rightarrow G_2$ be a homomorphism. Then

  1. $\phi (e) = e'$, where $e$ is the identity in $G_1$ and $e'$ is the identity in $G_2$.
  2. For all $g \in G , \phi (g^{-1}) = [\phi(g)]^{-1}$.
  3. $H_1 \leq G_1 \Rightarrow \phi (H_1) \leq G_2$.
  4. $H_2 \leq G_2$ (respectively, $H_2 \unlhd G_2$) $\Rightarrow \phi^{-1} (H_2) \leq G_1$ (respectively, $\phi^{-1} (H_2) \unlhd G_1$).

Theorem 9.4 Let $\phi : G \rightarrow H$ be a group homomorphism. Then the kernel
of $\phi$ is a normal subgroup of $G$.

Theorem 9.9 (first isomorphism thm) If $\phi G \rightarrow H$ is a homomorphism, then $G/ker\phi \cong \phi(G)$. In particular, if $\phi$ is onto, $G/ker\phi \cong \phi(H)$.

Theorem 9.11 (Correspondence Theorem) Let $N$ be a normal subgroup of a group $G$. Then $H \rightarrow H/N$ is a one -to-one correspondence between the set of subgroups $H$ containing $N$ and the set of subgroups of $G/H$. Furthermore, the normal subgroups of $H$ correspond to the normal subgroup of $G/N$.

Examples & Notes

Fact: $Ker \phi \unlhd G$.

Examples of homomorphisms: $\phi: S_n \longrightarrow \mathbb{Z}_2$, $\psi: Q_8 \longrightarrow V_4$

Example of Prop. 9.9 $\mathrm{ker}( \phi) =$ all multiples of $4 = \{\dots, -4,0,4,8, \dots\}= 4 \mathbb{Z}$

Example 1. Let $G$ be an abelian group. Every subgroup $H$ of $G$ is a
normal subgroup. Since $gh = hg$ for all $g \in G$ and $h \in H$, it will always be
the case that $gH = Hg$.

Example 2. Let $H$ be the subgroup of $S_3$ consisting of elements $(1)$ and
$(12)$. Since

\begin{align} (123)H = \{(123), (13)\} \end{align}


\begin{align} H(123) = \{(123), (23)\}, \end{align}

$H$ cannot be a normal subgroup of $S_3$. However, the subgroup $N$, consisting
of the permutations $(1)$, $(123)$, and $(132)$, is normal since the cosets of $N$are

\begin{align} N = \{(1), (123), (132)\} \end{align}
\begin{align} (12)N = N(12) = \{(12), (13), (23)\}. \end{align}

Example 3. Consider the normal subgroup of $S_3$,$N = \{(12),(123),(132)\}$. The cosets of $N$ in $S_3$ are $N$ and $(12)N$.

Example 5. Consider the dihedral group $D_n$, generated by the two elements
$r$ and $s$, satisfying the relations

\begin{align} r^{n} = id \\ s^{2} = id \\ srs = r^{-1}. \end{align}

The element $r$ actually generates the cyclic subgroup of rotations, $R_n$, of
$D_n$. Since $srs^{-1} = srs = r^{-1} \in R_n$, the group of rotations is a normal
subgroup of $D_n$; therefore, $D_n \diagup R_n$ is a group. Since there are exactly two
elements in this group, it must be isomorphic to $Z_2$.

Example 6. Let $G$ be a group and $g \in G$. Define a map $\phi : Z \rightarrow G$ by
$\phi (n) = g^{n}$. Then $\phi$ is a group homomorphism, since

\begin{align} \phi (m + n) = g^{m+n} = g^{m}g^{n} = \phi (m) \phi (n). \end{align}

This homomorphism maps $\mathbb{Z}$ onto the cyclic subgroup of $G$ generated by $g$.

Example 7. Let $G=GL_2(\mathbb{R})$. If

\begin{align} \[ A = \left( {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right) \] \end{align}

is in $G$, then the determinant is nonzero.
Example 8.
$\phi : S_3\rightarrow Z_2$

$\phi (\sigma)$ = {1, if odd and 0, if even}

$ker \phi = \{(1)(123)(132)\}$

$\{S_3: <(123)>\}=2$

Example 9. (Added for Jordan and Bradley)

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