Math4302 Modern Algebra (Lecture 2)

Recall from last lecture

Binary operations

A binary operation that is not associative but commutative:

Consider $(\mathbb{Z},*)$ where $a*b=|a-b|$ .

This is trivially commutative.

But $a=4,b=3,c=1$ gives $(a*b)*c=(4*3)*1=1*1=0$ . and $a*(b*c)=4*(3*1)=4*2=2$ .

Definition for identity element

An element $e\in X$ is called identity element if $a*e=e*a=a$ for all $a\in X$ .

Group

Definition of group

A group is a set $G$ with a binary operation $*$ that satisfies the following axioms:

Closure: $\forall a,b\in G, a* b\in G$ (automatically guaranteed by definition of binary operation).
Associativity: $\forall a,b,c\in G, (a* b)* c=a* (b* c)$ .
Identity element: $\exists e\in G, \forall a\in G, a* e=e* a=a$ .
Inverses: $\forall a\in G, \exists a^{-1}\in G, a* a^{-1}=a^{-1}* a=e$ .

Note

The inverse of $a$ is unique: If there is $b'\in G$ such that $b'*a=a*b'=e$ , then $b=b'$ .

Proof:

$b'=b'*e=b'*(a*b)=(b'*a)*b=e*b=b$ .

apply the definition of group.

Example of group

$(\mathbb{Z},+)$ is a group.

$(\mathbb{Q},+)$ is a group.

$(\mathbb{R},+)$ is a group.

with identity $0$ and all abelian groups.

$(\mathbb{Z},\cdot)$ , $\mathbb{Q},\cdot)$ , $(\mathbb{R},\cdot)$ are not groups ( $0$ has no inverse).

We can fix this by removing $0$ .

$(\mathbb{Q}\setminus\{0\},\cdot)$ , $(\mathbb{R}\setminus\{0\},\cdot)$ are groups.

$(\mathbb{Z}\setminus\{0\},\cdot)$ is not a group.

$(\mathbb{Z}_+,+)$ is not a group.

Consider $S$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ .

$(S,+)$

Identity: $f(x)=0$
Associativity: $(f+g)(x)=f(x)+g(x)$
Inverse: $f(x)=-f(x)$

This is a group.

$(S,\circ)$

Identity: $f(x)=x$
Associativity: $(f\circ g)(x)=f(g(x))$
Inverse: not all have inverse… (functions which are not bijective don’t have inverses)

This is not a group.

$\operatorname{GL}_(n,\mathbb{R})$ : set of $n\times n$ invertible matrices over $\mathbb{R}$ .

$(\operatorname{SL}_(n,\mathbb{R}),\cdot)$ where $\cdot$ is matrix multiplication.

Identity: $I_n$
Associativity: $(A\cdot B)\cdot C=A\cdot (B\cdot C)$
Inverse: $(A^{-1})^{-1}=A$

This is a group.

Matrix multiplication is not generally commutative, therefore it’s not abelian.

Definition of abelian group

A group $(G,*)$ is called abelian if $a* b=b* a$ for all $a,b\in G$ . ( $*$ is commutative)

Properties of group

$(a*b)^{-1}=b^{-1}* a^{-1}$

Proof

$(b^{-1}* a^{-1})*(a*b)=b^{-1}* a^{-1}*a*b=b^{-1}* e*b=b*b^{-1}=e$

$(a*b)* (b^{-1}* a^{-1})=a* b*b^{-1}* a^{-1}=a* e*a^{-1}=a*a^{-1}=e$

Cancellation from right and left:

a*b=a*c\implies b=c

b*a=c*a\implies b=c

Proof

\begin{aligned} a*b&=a*c\\ a^{-1}*(a*b)&=a^{-1}*(a*c)\\ e*b&=e*c\\ b&=c \end{aligned}

right cancellation are the same

Note

This also implies that every row/column of the table representation of the binary operation is distinct.

If not, suppose $a,b$ have the same row/column, then we can prove $a=b$ using cancellation from right and left.

We can solve equations $a*x=b \text{ and } x*a=b$ uniquely.

$x=a^{-1}* b$ , similarly $x=b* a^{-1}$ .

Finite groups

Group with 1 element $\{e\}$ .

Group with 2 elements $\{e,a\}$ . (example is $(\{-1,1\},\times)$ )

And

*	e	a
e	e	a
a	a	e

Group with 3 elements $\{e,a,b\}$ .

And the possible ways to fill the table are:

*	e	a	b
e	e	a	b
a	a	b	e
b	b	e	a