Algebraic Structures Practical 1

Question 1.1

1. Set: $\{1\}$ with multiplication,

   • Closure: $1\cdot 1=1\in \{1\}$.
   • Associativity: $1\cdot 1=1=1\cdot 1$.
   • Identity Element: $1$.
   • Inverse Element: $1$ is its own inverse.
   • Conclusion: This is a (trivial) group.

2. Set: $\mathbb{C}$ with addition,

   • Closure: $(a+ib)+(c+id)=(a+c)+i(b+d)\in \mathbb{C}$.
   • Associativity: $((a+ib)+(c+id))+(e+if)=(a+c+e)+i(b+d+f)=(a+ib)+((c+id)+(e+if))$.
   • Identity Element: $0+0i$ (the complex number $0$).
   • Inverse Element: For any $a+ib$, the inverse is $-a-ib$.
   • Conclusion: This is a group.

3. Set: $\mathbb{C}$ with multiplication,

   • Closure: $(a+ib)(c+id)=(ac-bd)+(ad+bc)i\in \mathbb{C}$.
   • Associativity: $((a+ib)(c+id))(e+if)=(ac-bd)e+(ad+bc)i+...$ (follows from the distributive property).
   • Identity Element: $1+0i$ (the complex number $1$).
   • Inverse Element: For any $a+ib\ne 0$, the inverse is $\frac{a-ib}{a^2+b^2}$.
   • Conclusion: This is a group, excl. zero

4. Set: $\{1,-1\}$ with multiplication,

   • Closure: $1\cdot 1=1$, $1\cdot (-1)=-1$, $(-1)\cdot 1=-1$, $(-1)\cdot (-1)=1$.
   • Associativity: $(x\cdot y)\cdot z=x\cdot (y\cdot z)$ for all $x,y,z\in \{1,-1\}$.
   • Identity Element: $1$.
   • Inverse Element: Each element is its own inverse: $1^{-1}=1$, $(-1{)}^{-1}=-1$.
   • Conclusion: This is a group.

5. Set: $\{-1,0,1\}$ with addition,

   • Closure: $(-1)+(-1)=-2\notin \{-1,0,1\}$; thus, not closed.
   • Associativity: Addition is associative in general.
   • Identity Element: $0$.
   • Inverse Element: $1^{-1}=-1$, $0^{-1}=0$, $(-1{)}^{-1}=1$.
   • Conclusion: This is not a group, as it’s not closed.

6. Set: $\mathbb{Z}$ with operation defined by $a\odot b=a+b-3$,

   • Closure: For any $a,b\in \mathbb{Z}$, $a\odot b=a+b-3\in \mathbb{Z}$.
   • Associativity: $a\odot (b\odot c)=a+(b+c-3)-3=a+b+c-6$ and $(a\odot b)\odot c=(a+b-3)+c-3=a+b+c-6$.
   • Identity Element: The identity element $e$ satisfies $a\odot e=a+e-3\Rightarrow e=3$.
   • Inverse Element: For $a\in \mathbb{Z}$, the inverse is $b$ such that $a\odot b=3\Rightarrow b=3-a$.
   • Conclusion: This is a group.

7. Set: $\mathbb{R}$ with operation defined by $a\ast b=a+b-ab$,

   • Closure: For any $a,b\in \mathbb{R}$, $a\ast b=a+b-ab\in \mathbb{R}$.
   • Associativity: $a\ast (b\ast c)=a+(b+c-bc)-a(b+c-ab)=...$ (requires verification).
   • Identity Element: The identity element $e$ satisfies $a\ast e=a+e-ae\Rightarrow e=1$.
   • Inverse Element: For $a\in \mathbb{R}$, the inverse is $b$ such that $a\ast b=1\Rightarrow b=\frac{1-a}{a}$ (for $a\ne 1$).
   • Conclusion: This is a group, excl. one

8. Set: $\mathbb{Z}[x]$ of polynomials with integer coefficients with multiplication.

   • Closure: The product of two polynomials $f(x)$ and $g(x)$ is a polynomial with integer coefficients, hence $f(x)g(x)\in \mathbb{Z}[x]$.
   • Associativity: Polynomial multiplication is associative: $(f\cdot g)\cdot h=f\cdot (g\cdot h)$ for any polynomials $f,g,h\in \mathbb{Z}[x]$.
   • Identity Element: The identity element is the polynomial $1$, since $f(x)\cdot 1=f(x)$ for any polynomial $f(x)$.
   • Inverse Element: There are no inverses in $\mathbb{Z}[x]$ for polynomials of degree greater than 0, as the product of two non-constant polynomials cannot yield the identity polynomial $1$.
   • Conclusion: This is not a group, as there are no inverses for non-constant polynomials.

9. Set: $\mathbb{Q}[x]$ of polynomials with rational coefficients with addition.

   • Closure: The sum of two polynomials $f(x)$ and $g(x)$ with rational coefficients is also a polynomial with rational coefficients, hence $f(x)+g(x)\in \mathbb{Q}[x]$.
   • Associativity: Polynomial addition is associative: $(f+g)+h=f+(g+h)$ for any polynomials $f,g,h\in \mathbb{Q}[x]$.
   • Identity Element: The identity element is the polynomial $0$, since $f(x)+0=f(x)$ for any polynomial $f(x)$.
   • Inverse Element: For any polynomial $f(x)$, the inverse is $-f(x)$, since $f(x)+(-f(x))=0$.
   • Conclusion: This is a group.

Question 1.2

If $G$ is a group operation denoted multiplicatively, then $(gh{)}^{-1}=h^{-1}{g}^{-1}$, as shown:

…

Question 1.3

Consider the set of diagonal matrices…

$$D=\left\{\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]:a,b\in {\mathbb{R}}^{\ast }\right\}$$

Part A

$D$ with usual matrix multiplication is a group:

• Closure: …
• Associativity: …
• Identity Element: …
• Inverse Element: …

Part B

This group is/isn’t abelian because …