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179 lines
3.7 KiB
Markdown
179 lines
3.7 KiB
Markdown
# Math4302 Modern Algebra (Lecture 2)
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## Recall from last lecture
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### Binary operations
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A binary operation that is not associative but commutative:
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Consider $(\mathbb{Z},*)$ where $a*b=|a-b|$.
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This is trivially commutative.
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But $a=4,b=3,c=1$ gives $(a*b)*c=(4*3)*1=1*1=0$. and $a*(b*c)=4*(3*1)=4*2=2$.
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#### Definition for identity element
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An element $e\in X$ is called identity element if $a*e=e*a=a$ for all $a\in X$.
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### Group
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#### Definition of group
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A group is a set $G$ with a binary operation $*$ that satisfies the following axioms:
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1. Closure: $\forall a,b\in G, a* b\in G$ (automatically guaranteed by definition of binary operation).
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2. Associativity: $\forall a,b,c\in G, (a* b)* c=a* (b* c)$.
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3. Identity element: $\exists e\in G, \forall a\in G, a* e=e* a=a$.
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4. Inverses: $\forall a\in G, \exists a^{-1}\in G, a* a^{-1}=a^{-1}* a=e$.
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> [!NOTE]
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>
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> The inverse of $a$ is unique: If there is $b'\in G$ such that $b'*a=a*b'=e$, then $b=b'$.
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>
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> Proof:
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>
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> $b'=b'*e=b'*(a*b)=(b'*a)*b=e*b=b$.
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>
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> apply the definition of group.
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<details>
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<summary>Example of group</summary>
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$(\mathbb{Z},+)$ is a group.
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$(\mathbb{Q},+)$ is a group.
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$(\mathbb{R},+)$ is a group.
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with identity $0$ and all abelian groups.
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---
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$(\mathbb{Z},\cdot)$, $\mathbb{Q},\cdot)$, $(\mathbb{R},\cdot)$ are not groups ($0$ has no inverse).
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---
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We can fix this by removing $0$.
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$(\mathbb{Q}\setminus\{0\},\cdot)$, $(\mathbb{R}\setminus\{0\},\cdot)$ are groups.
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---
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$(\mathbb{Z}\setminus\{0\},\cdot)$ is not a group.
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$(\mathbb{Z}_+,+)$ is not a group.
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---
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Consider $S$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.
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$(S,+)$
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- Identity: $f(x)=0$
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- Associativity: $(f+g)(x)=f(x)+g(x)$
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- Inverse: $f(x)=-f(x)$
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This is a group.
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$(S,\circ)$
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- Identity: $f(x)=x$
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- Associativity: $(f\circ g)(x)=f(g(x))$
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- Inverse: not all have inverse...... (functions which are not bijective don't have inverses)
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This is not a group.
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---
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$\operatorname{GL}_(n,\mathbb{R})$: set of $n\times n$ invertible matrices over $\mathbb{R}$.
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$(\operatorname{SL}_(n,\mathbb{R}),\cdot)$ where $\cdot$ is matrix multiplication.
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- Identity: $I_n$
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- Associativity: $(A\cdot B)\cdot C=A\cdot (B\cdot C)$
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- Inverse: $(A^{-1})^{-1}=A$
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This is a group.
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**Matrix multiplication is not generally commutative**, therefore it's not abelian.
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</details>
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#### Definition of abelian group
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A group $(G,*)$ is called abelian if $a* b=b* a$ for all $a,b\in G$. ($*$ is commutative)
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#### Properties of group
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1. $(a*b)^{-1}=b^{-1}* a^{-1}$
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<details>
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<summary>Proof</summary>
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$(b^{-1}* a^{-1})*(a*b)=b^{-1}* a^{-1}*a*b=b^{-1}* e*b=b*b^{-1}=e$
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$(a*b)* (b^{-1}* a^{-1})=a* b*b^{-1}* a^{-1}=a* e*a^{-1}=a*a^{-1}=e$
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</details>
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2. Cancellation from right and left:
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$$
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a*b=a*c\implies b=c
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$$
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$$
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b*a=c*a\implies b=c
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$$
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<details>
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<summary>Proof</summary>
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$$
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\begin{aligned}
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a*b&=a*c\\
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a^{-1}*(a*b)&=a^{-1}*(a*c)\\
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e*b&=e*c\\
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b&=c
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\end{aligned}
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$$
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right cancellation are the same
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</details>
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> [!NOTE]
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>
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> This also implies that every row/column of the table representation of the binary operation is distinct.
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>
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> _If not, suppose $a,b$ have the same row/column, then we can prove $a=b$ using cancellation from right and left._
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3. We can solve equations $a*x=b \text{ and } x*a=b
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$ uniquely.
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$x=a^{-1}* b$, similarly $x=b* a^{-1}$.
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### Finite groups
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Group with 1 element $\{e\}$.
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Group with 2 elements $\{e,a\}$. (example is $(\{-1,1\},\times)$)
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And
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|*|e|a|
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|-|-|-|
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|e|e|a|
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|a|a|e|
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Group with 3 elements $\{e,a,b\}$.
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And the possible ways to fill the table are:
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|*|e|a|b|
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|-|-|-|-|
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|e|e|a|b|
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|a|a|b|e|
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|b|b|e|a|
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