Examples

• \(\mathbb{N}=\{1,2,3,...\}\) : they exist naturally

• \(\mathbb{Z}=\{...,-3,-2,-1,0,1,2,3,...\}\) : includes zero and negatives

• \(\mathbb{Q}=\{\frac{m}{n}|m,n\in \mathbb{Z}\text{ and }n\ne 0\}\) : integer fractions

• \(\mathbb{R}\) : includes square roots and pie, real analysis starts with \(\sqrt{2}\)

• \(\mathbb{C}\) : includes imaginary numbers

• Asterisk in the superscript means delete zero

• Plus sign in the superscript means only positive values (>0)

Properties

1. * is commutative if \(a\ne b\), \(a*b=b*a\) \(\forall\) a,b \(\in A\).

   • + and \(\cdot\) are commutative

   • - , \(\div\) , funtion composition and matrix multiplication are not commutative

2. * is associative if \((a*b)*c=a*(b*c)\) \(\forall\) a,b,c, \(\in A\).

   • addition is associative, subtraction is not associative

3. If \(\exists\) \(e\in A\) \(\Rightarrow\) \(e*a=a*e=a\) \(\forall\) \(a\in A\), the we call e the identity element in A w.r.t. *.

   • 0=e w.r.t. addition

   • 1=e w.r.t. multiplication

4. If \(e\in A\) is the identity w.r.t. * and \(a,b\in A\Rightarrow\) \(a*b=b*a=e\) we call a and b inverses of one another.

   • the inverse of \(a\in \mathbb{R}\) w.r.t. addition is -a since \(a+(-a)=(-a)+a=0\)

   • the inverse of \(a\in\mathbb{R}^*\) w.r.t. multiplication is \(\frac{1}{a}\) since \(a(\frac{1}{a})=(\frac{1}{a})a=1\)

Proof Outlines

1. Commutative:

   \(\underline{\text{No}}\): Give an example, “Let a = 1, and b =2”, and then show \(a*b\ne b*a\).

   \(\underline{\text{Yes}}\): “For any a,b in the set” and then show \(a*b=b*a\).

2. Associative

   \(\underline{\text{No}}\): Give an example, “Let a=1, b=2, c=3” and then show \(a*(b*c)\ne (a*b)*c\).

   \(\underline{\text{Yes}}\): “For any a,b,c in the set” and then show \(a*(b*c)=(a*b)*c\).

3. Identity

   \(\underline{\text{No}}\): Suppose that \(e\in\) the given set \(\Rightarrow\) \(a*e=a\) \(\forall a\in\) the given set. Then show \(a*e=a\) by plugging e in for b and solving for e. “Since the identity element must be a constant then there is no identiy w.r.t.” the given set. (can’t involve variables)

   \(\underline{\text{Yes}}\): State what the identity element is w.r.t. the orperation and show that \(a*e=a\) and \(e*a=a\).

4. Inverses

   \(\underline{\text{No}}\): Given an example of an element who doesn’t have an inverse.

   Note: If there is not identity element then there is no inverse.

   \(\underline{\text{Yes}}\): “Suppose b=\(a^{-1}\). Then \(a*b=e\)” (where e is the identity found in 3), and then try to solve the eqation for b. Then check \(b*a=e\) as well.

   Note: Do not need to check \(b*a=e\) if we know * is commutative.

Proposition 1

Let G be a group, then G has exactly one idently element.

Proposition 2

Every element of G has exactly one inverse.

Theorem 1 (Cancellation Law)

Let G be a group and let \(a,b,c\in G\), then \(ab=ac\Rightarrow b=c\) and \(ba=ca\Rightarrow b=c\).

Theorem 2

Let G be a group and let \(a,b\in G\). If \(ab=e\), then a and b are inverses, i.e. \(a=b^{-1}\) and \(b=a^{-1}\).

Theorem 3

Let G be a group and let \(a,b\in G\) then \((ab)^{-1}=b^{-1}a^{-1}\) and \((a^{-1})^{-1}=a\).

• to show a and b are inverses, show their product is e.

Example

Addition: i) \(0\in H\). ii) \(\forall\) a,b \(\in H\) \(a+b=H\). iii) \(\forall\) \(a\in H\) , \(-a\in H\).

Proof Outlines (Two-Step Subgroup Test)

1. \(e\in H\)

2. for all a,b \(\in H\), \(ab^{-1}\in H\).

Prove something is a subgroup of G.

1. “Suppose” then show e is 0 or 1 for the subgroup in a short series of equalities, “the additive or multiplicative element 0 or 1 is in” the subgroup.

2. “Now take any a,b \(\in\) the subgroup.” Then define a and b potentially for some other integers. Then show \(ab^{-1}=\) something identifiably in the subgroup.

“Therefore we’ve shown that” our subgroup “contains the identity element and for all a,b \(\in H\) , \(ab^{-1}\in\)” our subgroup. Thus our subgroup is a subgroup of G.

Added Notes

1. If G is abelian, then \((ab)^n=a^nb^n\) for all \(n\in \mathbb{N}\).

2. In any group \((a^{-1})^n=(a^n)^{-1}\) for all \(n\in \mathbb{N}\).

3. In any group \(e^{-1}=e\).

Two-step subgroup test

Let G be a group. A subset H \(\subseteq G\) is a subgroup of G if

1. \(e\in H\)

• addition: show 0 \(\in H\)

2. \(\forall a,b,\in H\), \(ab^{-1}\in H\).

• addition: \(\forall\) a,b \(\in H\), \(a+(-b)=a-b\in H\)

Injective

A function \(f:A\rightarrow B\) is injective or one-to-one if \(f(x_1)=f(x_2)\) \(\Rightarrow\) \(x_1=x_2\).

Surjective

A function \(f:A\rightarrow B\) is surjectie or onto if for all \(y\in B\) there exists \(x\in A\) such that \(f(x)=y\).

Bijective

A function \(f:A\Rightarrow B\) is bijectie because f is both one-to-one and onto.

Proof Outline : Bijection, proving injective and surjective

1. Injective

“Suppose for some \(x_1, x_2\in G\),” then show \(f(x_1)=f(x_2)\) \(\Rightarrow\) \(x_1=x_2\). “So, f is one-to-one”.

2. Surjective

“Take any y \(\in G\). We need to find \(x\in G\Rightarrow f(x)=y\).” Then solve for x, and check that the x you found works for \(f(x)=y\). “Therefore f is also onto”.

“We’ve shown that \(f\) is injective and surjective. So by definition, f is a bijection from G to G”.

Cycle Decomposition

Example: \(\Rightarrow f=(1523)(47)\)

Disjoint Cyccle

We call two cycles disjoint if they have no numbers in common.

Disjoint cycles in \(S_n\) commute.

Inverse of a Cycle

Example: \(f^{-1}=(\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 3 & 5 & 2 & 7 & 1 & 6 & 4\end{smallmatrix})\)

Length and Transposition

The number of integers in a cycle is called its length.

A cycle of length 2 is called a transposition.

Isomorphism Proposition 1

Let f: \(G_1\rightarrow G_2\) be an isomorphism. Then

1. \(f(e_1)=e_2\)

2. \(\forall a\in G\), f(a) and \(f(a)^{-1}\) are inverese in \(G_2\), i.e. \(f(a^{-1})=[f(a)]^{-1}\)

3. If \(G_1=\langle a\rangle\), then \(G_2=\langle f(a)\rangle\).

\(\cong\) Groups have the same algebraic properties

Suppose \(G_1\cong G_2\). Then,

1. \(|G_1|=|G_2|\) (they have the same size)

2. \(G_1\) is abelian \(\Leftrightarrow\) \(G_2\) is abelian.

3. \(G_1\) is cyclic \(\Leftrightarrow\) \(G_2\) is cyclic.

4. \(G_1\) has the property that every element squared equals e \(\Leftrightarrow\) \(G_2\) has the same property.

5. Every subgroup of \(G_1\) is isomorphic to a subgroup of \(G_2\).

and so on…

Groups Size 4

Every group of size 4 is isomorphic to \(\mathbb{Z_}_4\) or \(\mathbb{Z}_2\times\mathbb{Z}_2\). (We proved that those two are not isomorphic to each other)