NoteNextra-origin/content/Math4302/Math4302_L2.md

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:

1. Closure: \forall a,b\in G, a* b\in G (automatically guaranteed by definition of binary operation).
2. Associativity: \forall a,b,c\in G, (a* b)* c=a* (b* c).
3. Identity element: \exists e\in G, \forall a\in G, a* e=e* a=a.
4. 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.

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

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We can fix this by removing 0.

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

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

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

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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.

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\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

1. (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

2. 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.

3. We can solve equations $ax=b \text{ and } xa=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