chapter 2 summary


A rearrangement of a geometric figure preserving the arrangement of its sides and vertices as well as its distances and angles.

Rigid Motion
A map from a plane to itself while keeping symmetry of the object.

Binary Operations
A binary operation or law of composition on a set $G$ is a function $G \times G \mapsto G$ that assigns to each pair $(a,b) \in G$ a unique element $a \circ b$, or $ab$ in $G$, called the composition of $a$ and $b$.

A group $(G, \circ)$ is a set $G$ together with a binary operation $\circ:(a,b) \mapsto a \circ b$ that satisfies:

  1. $\circ$ is associative: $(a\circ b) \circ c=a\circ (b\circ c); \forall a,b,c\in G$.
  2. There exists an identity, denoted by $e$ (or $0$ or $1$): $a\circ e = e\circ a = a; \forall a\in G$.
  3. For each $a \in G$, there exists an inverse, denoted by $a^{-1}$: $a\circ a^{-1}$=$a^{-1}\circ a$=$e$.

Identity Element
an element $e\in G$, such that for any element $a\in G$ , $a\circ e = e\circ a = a$

Abelian Groups
A group $G$ with the property that $a\circ b = b\circ a$ for all $a,b\in G$ is called abelian or commutative. Groups not satisfying this property are said to be nonabelian or noncummutative.

Non-Abelian Group
$S_3$ is an example of a 6 element group that is not abelian. $3!$
$V_4$ is a non-abelian group that is the symmetry group for a non-square rectangle: $\{e, h, v, r\}$

A subgroup $H$ of a group $G$ is a subset of $G$ such that the group operation restricted to $H$ ($H$ has the same operation as $G$), $H$ is a subgroup in its own right.

Proper Subgroup
If $H\leq G$ and $H\neq G$, then H is called a Proper Subgroup.

Finite of Finite Order
A group is finite, or has finite order, if it contains a finite number of elements. If the group is not finite then it is said to be infinite or have infinite order.

The order of a finite group is the number of elements that it contains. The group $G$ containing $n$ elements is written $\mid G \mid = n$.
Example- The group $\mathbb{Z}_5$ is a finite group of order 5.


Proposition 2.1
Let $n$ be a set of equivalence classes in the integers $mod n$ and $a,b,c\in\mathbb{Z}$.
1. Addition and multiplication can be commutative:
$a+b$ $\equiv$ $b+a$ mod $n$
$ab$ $\equiv$ $ba$ mod $n$
2. Addition and Multiplication can be associative:
$(a+b)+c$ $\equiv$ $a+(b+c)$ mod $n$
$(a*b)$ $c$ $\equiv$ $a$ $(b*c)$
3. The set has an additive identity and multiplicative identity:
$a+0$ $\equiv$ $a$ mod $n$
$a*1$ $\equiv$ $a$ mod $n$
4. Multiplication can distribute over addition:
$a$$(b+c)$ $\equiv$ $a*b+a*c$ mod $n$
5. For every integer $a$ there is an additive inverse $-a$:
$a+(-a)$$\equiv$$0$ mod $n$

Proposition 2.2
The identity element in a group $G$ is unique.

Proof. Suppose that $e$ and $e^{l}$ are both identities in G. Then $eg = ge = g$
and $e^{l}g = ge0 = g$ for all $g\in G$. We need to show that $e = e^{l}$. If we think
of $e$ as the identity, then $ee^{l} = e^{l}$; but if $e^{l}$ is the identity, then $ee^{l} = e$.
Combining these two equations, we have $e = ee^{l} = e^{l}$.

inverses in a group are also unique. If $g^{l}$and $g^{ll}$ are both inverses of an
element $g$ in a group G, then $gg^{l} = g^{l}g = e$ and $gg^{ll} = g^{ll}g = e$. We want
to show that $g^{l} = g^{ll}$, but $g^{l} = g^{l}e = g^{l}(gg^{ll}) = (g^{l}g)g^{ll} = eg^{ll} = g^{ll}$. We
summarize this fact in the following proposition.

Proposition 2.3
Let $g\in$$G$, then the inverse of g is unique, $g^{-1}$

Proposition 2.4
Let $G$ be a group and $a,b\in$$G$, we see that

Proposition 2.5
Let $G$ be a group. For any $a\in G$, $(a^{-1})^{-1}=a$.

Proposition 2.6
Let $G$ be a group and $(a,b)\in$$G$, then
$ax=b$ and $xa=b$ have unique solutions in $G$.

Proposition 2.7 Cancellation Law for Groups
If $G$ is a group and $a,b \in G$ then $ba=ca$ or $ab=ac$ implies $b=c$.

Theorem 2.8
If $G$ is a group and $g\in$$G$, then we define $g^0=e$. For $n\in$$N$,
we define

\begin{align} g^n=\underbrace{gg\cdots g}_{n\text{ times}} \end{align}

In a group, the usual laws of exponents hold; that is, for all $g,h\in\G$,
1. $g^m g^n = g^{m+n}$ for all $m,n\in\mathbb{Z}$;
2. $(g^m)^n = g^{mn}$ for all $m,n\in\mathbb{Z}$;
3. $(gh)^n = (h^{-1} g^{-1})^{-n}$ for all $n\in\mathbb{Z}$. Furthermore, if G is abelian, then $(gh)^n = g^n h^n$.

Proposition 2.9
A subset $H$ of $G$ is a subgroup iff it satisfies the following conditions:
1. The identity $e$ of $G$ in $H$.
2. If $h_1, H_2 \in H$, then $h_1h_2 \in H$.
3. If $h \in H$, then $h^{-1} \in H$.

Proposition 2.10
Let $H$ be a subset of a group $G$. Then $H$ is a subgroup of $G$ iff $H\neq \emptyset$, and whenever $g,h\in H$, $gh^{-1}\in H$.

Examples & Notes

Exercise 24
Let a and b be elements in a group $G$; Prove that $(ab)^{n}(a)^{-1} = (aba^{-1})^{n}$

The statement for n = 1 is simply that $aba^{-1}= a b a^{-1}$, which is certainly true. Now assume that the result holds for $n = k$.

$(aba^{-1})^{k+1} = (aba^{-1})^{k}(a b a^{-1})$
$(a b^{k} a^{-1}) (a b a^{-1})$
$(a b^{k})(a^{-1} a)( b a^{-1} )$
$(a b^{k})(b a^{-1})$
$a b^{k+1} a^{-1}$

Therefore the statement is true for $n=k+1$ which means it holds true for all values of n.

Exercise 28
Prove the left and right cancellation laws for a group $G$; that is, show that in the group $G$, $ba = ca$ implies $b = c$ and $ab = ac$ implies $b = c$ for elements $a,b,c\in G$.

Let $a,b,c\in G$, and assume $ba = ca$ and $ab = ac$.
We see that:

\begin{align} ba&= ca\\ ba(a^{-1})&= ca(a^{-1})\\ be&= ce\\ b&= c \end{align}


\begin{align} ab&=ca\\ (a^{-1})ab&= (a^{-1})ac\\ eb &= ec\\ b&= c. \end{align}

Claim: The collection of symmetries of an object forms a group under composition (means doing one symmetry and than another). Example $S_{3} \cong D_{3}$
Important: Function composition is associative so composing permutations is associative.


\begin{align} \begin{flushleft} e ={\left( \begin{array}{ccc} A & B & C \\ A & B & C \end{array} \right)} r_1= {\left( \begin{array}{ccc} A & B & C \\ C & A & B \end{array} \right)} r_2= {\left( \begin{array}{ccc} A & B & C \\ B & C & A \end{array} \right)} f_1= {\left( \begin{array}{ccc} A & B & C \\ A & C & B \end{array} \right)} f_2={\left( \begin{array}{ccc} A & B & C \\ B & A & C \end{array} \right)} f_3= {\left( \begin{array}{ccc} A & B & C \\ C & B & A \end{array} \right)} \end{align}

Some Examples of Groups
1. $\mathbb{Z}$ under +, is a abelian group
2. $\mathbb{Z}$ under multiplication, is not a group, fails inverses
3. $\mathbb{R}$ under + is a abelian group
4. $\mathbb{R}$ under multiplication, is not a group, 0 fails to have an inverse
5. $\mathbb{R*} =\mathbb{R} - {0}$ is a abelian group under multiplication
6. $\mathbb{Z}_{n}$ is a abelian group under + mod $n$
7. $\mathbb{C*} = \mathbb{C} - {0{$ is a group under multiplication
8. $GL_{2} =$ set of all $2 \times 2$ invertible matrices, is a group under matrix multiplication

Exercise #29

Show that if $a^{2} = e$ $\forall$ $a \in G$ then G must be an abelian group.

Proof: Let $a,b \in G$ and $a^{2} = e$
Assume: $g^{2} = e$ $\forall$ $g\in G$

We see that:

$abab = (ab)^{2}=e$
$abab = e$
$aabab = ae$
$ebab = a$
$bab = a$
$bbab = ba$
$eab = ba$
$ab = ba$

Proposition 2.8 pt3 abelian Proof
Given $G$ is a group, $(ab)^2 = a^2b^2$ for all $a,b \in G$ if $G$ abelian.
We know:
$(ab)^2 = a^2b^2$
This means:
$abab = aabb$
Multiply each side on left by $a^{-1}$
$a^{-1}abab =a^{-1}aabb$
The inverse cancels the first $a$'s
$bab = abb$
Now multiply on right by $b^{-1}$
$babb^{-1} = abbb^{-1}$
The right $b's$ cancel
Therefore, group $G$ is abelian!!

Cayley Table for $\mathbb{Z}_{4}$

Unsupported math environment "tabular"

Subgroup lattice of $Q_8$ (sorry I didn't know how to draw the lines)

Unsupported math environment "center"

subgroups $D_3$
$\{{e}\}$, $\{{e,f_3}\}$, $\{{e,f_1}\}$, $\{{e,f_2}\}$, $\{{e, r_1, r_2}\}$, $D_3$
where f represents each of the different flip and $r_1$ is a 120 degree rotation and $r_2$ is a 240 rotation.

ij = k

jk = i

ki = j

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