chapter 8 summary

# Definitions

Definitions go here.

Isomorphic
Two groups $(G , \centerdot)$ and $(H , \circ)$ are isomorphic if $\exists$ a 1-1 and onto map $\phi : G \rightarrow H$ such that the group operations are preserved.

(1)
\begin{align} \phi(a \cdot b) = \phi(a) \circ \phi(b) \end{align}

for all $a,b \in G$. We write $G \cong H$.

# Theorems

Theorems, propositions, corollaries, and lemmas go here.

Theorem 8.1
Let $\phi : G \rightarrow H$ be an isomorphic. Then the following are true.

1. $\phi^{-1} : H \rightarrow G$ is also isomorphic.
2. $|G| = |H|$
3. If $G$ is abelian, then $H$ is abelian.
4. If $G$ is cyclic, then $H$ is cyclic.
5. If $G$ has a subgroup of order $n$, then $H$ has a subgroup of order $n$.

Theorem 8.2
All cyclic groups of infinite order are isomorphic to $\mathbb Z$
Pf.
Let $G$ be a cyclic group of infinite order and suppose that $a$ is a generator of $G$. Define a map $\phi: \mathbb Z \rightarrow G \text{ by} \phi : n\mapsto a^n.$ Then

(2)
\begin{align} \phi{(m+n)}=a^{m+n}=a^m a^n=\phi{(m)}\phi{(n)}. \end{align}

To show that $\phi$is injective, suppose that $m$ and $n$ are two elements in $\mathbb Z$,
where $m\neq n$. We can assume that $m> n$. We must show that $a^m \neq a^n$.
Let us suppose the contrary; that is, $a^m = a^n$. In this case $a^{m-n} = e$, where
${m-n} > 0$, which contradicts the fact that $a$ has infinite order. Our map
is onto since any element in $G$ can be written as $a^n$ for some integer $n$ and
$\phi{(n)} = a^n$.

Theorem 8.3
If $G$ is a cyclic group of order $n$, then $G \cong \mathbb{Z}_n$.

Corollary 8.4
If $G$ is a group of order $p$, where $p$ is prime, then $G \cong \mathbb{Z}_p$.

Theorem 8.5
The isomorphism of groups is an equivalence relation (i.e. a partition) of the class of groups.

Theorem 8.7
Let $(G, \cdot)$ and $(H, \circ)$be two groups. Then

(3)
\begin{align} G \times H = \{(g,h): g \in G, h \in H\} \end{align}

is a group under the operation

(4)
\begin{align} (g_1,h_1g_2,h_2) = (g_1 \cdot g_2, h_1 \circ h_2). \end{align}

Note: Would need to check identity, inverse & closure. So the
(i) identity: $(e_G, e_H)$, where $e_G$is identity in $G$and $e_H$is identity in $H$.
(ii)inverse: $(g_1 h)^{-1} = (g^{-1}, h^{-1})$.

Would be great to have some examples here…

Theorem 8.8
Let $g, h\in G \times H$ If $|g|=k$ , and $|h|=m$, then $|(g,h)|=LCM(k,m)$ .

Theorem 8.10
The group $\mathbb Z_m \times \mathbb Z_n \cong \mathbb Z_m_n$ iff $gcd(m,n)= 1$, relatively prime

# Examples & Notes

If two Cayley tables/diagrams are the same "up to relabeling", then groups are isomorphic.
(Note: The converse is not true.)

$"=" \Rightarrow "\cong"$. It is not true the other way around.

Example 5. Even though $\mathbb S_3$ and $\mathbb Z_6$ possess the same number of elements,
we would suspect that they are not isomorphic, because $\mathbb Z_6$ is abelian and
$\mathbb S_3$ is nonabelian. To demonstrate that this is indeed the case, suppose that
$\phi : \mathbb Z_6 \rightarrow \mathbb S_3$ is an isomorphism. Let $a, b \in \mathbb S_3$ be two elements such that
$ab \neq ba$. Since $\phi$ is an isomorphism, there exist elements $m$ and $n$ in $\mathbb Z_6$
such that
$\phi (m) = a$
$\phi (n) = b$.
However,
$ab = \phi (m)\phi (n) = \phi(m + n) = \phi(n + m) = \phi(n)\phi (m) = ba$,
which contradicts the fact that $a$ and $b$ do not commute.

Example Isomorphic Proof
$\mathbb Z_2 = \{0,1\}$, $H=<S> = \{e,s\}$
$\mathbb Z_2 \cong H$
Proof:
Let $\phi : \mathbb Z_2 \rightarrow H$ be defined by $\phi (0) = e$ and $\phi (1)= s$ then $\phi$ is one to one, and onto.
Let a,b $\in \mathbb Z_2$. Then $\phi (a+b)= S^{a+b}=S^{a} S^{b} = \phi (a) \phi (b)$.
So, $\phi$ is isomorphic.

$Z_2 \times Z_2$ is not iso to $Z_4$

$Z_2 \times Z_2$ is not cyclic, $Z_4$ is cyclic

$Z_2 \times Z_2$ is iso to $V_4$

Ex. 8.27
Let $G \cong H$. Show that if $G$ is cyclic, then so is $H$.

Proof.
Assume that $G \cong H$ and $G$ is cyclic. We need to show $H$ is cyclic.

Since $G \cong H$, then $\exsits f: G \rightarrow H$ s.t. $f$ is an isomorphism.

We see that if $G$ is cyclic, $\exists$ $a \in G$s.t. $\langle a \rangle = G$.

Let $x \in H$. Since $f$ is onto, $\exists$ $b \in G$s.t. $f(b) = x$. Then, $\exists$ $k$ s.t. $b = a^k$.

So, $x = f(b) = f(a^k)$

$= \underbrace{f(a) \dots f(a)}_{k} = f(a)$. $QED$.

Facts

1. $\phi(e) = e'$, $e \in G$, $e' \in H$
2. $\phi(a^{-1}) = (\phi(a) )^{-1}$

Does not require that one-to-one or onto for above facts to be true.

Example of Theorem 8.8
In $S_3 \times \mathbb Z_3$
$|(s,1)|= LCM(|s|, |1|)= LCM (2,3) = 6$

Example of Theorem 8.10
(a). $\mathbb Z_2 \times \mathbb Z_2 \cong \mathbb Z_4$?
Not Isomorphic: $gcd(2,2) = 2\ne 1$

(b). $\mathbb Z_2 \times \mathbb Z_3 \cong \mathbb Z_6$?
Since the $gcd(2,3) =1$, then its isomorphic.

page revision: 30, last edited: 18 May 2010 03:09