11 Basic Axioms and Examples (Groups)

#todo/school : Do problems 1ad, 2, 6abcd, 9, 11, 12, 25 (pp. 21 (35)) 📅 2025-01-08 ✅ 2025-01-08

1ad

Question

Determine which of the following binary operations are associative:
a. the operation $⋆$ on $Z$ defined by $a ⋆ b = a - b$
b. the operation $⋆$ on $R$ defined by $a ⋆ b = a + b + a b$
c. the operation $⋆$ on $Q$ defined by $a ⋆ b = \frac{a + b}{5}$
d. the operation $⋆$ on $Z \times Z$ defined by $(a, b) ⋆ (c, d) = (a d + b c, b d)$
e. the operation $⋆$ on $Q - {0}$ defined by $a ⋆ b = \frac{a}{b}$

Proof

\begin{aligned} (a ⋆ b) ⋆ c & = (a - b) ⋆ c \\ = (a - b) - c \\ \neq a - c + b \\ = a - (b - c) \\ = a ⋆ (b ⋆ c) \end{aligned}

So we just need an example where $(a - b) - c \neq a - (b - c) \Rightarrow c \neq - c$ so choose $a = 1, b = 1, c = 1$ as a counterexample. Thus this isn't associative.

\begin{aligned} ((a, b) ⋆ (c, d)) ⋆ (e, f) & = (a d + b c, b d) ⋆ (e, f) \\ = ((a d + b c) f + b d e, b d f) \\ = (a d f + b c f + b d e, b d f) \\ = (a d f + b (c f + d e), b d f) \\ = (a, b) ⋆ (c f + d e, d f) \\ = (a, b) ⋆ ((c, d) ⋆ (e, f)) \end{aligned}

So it is associative!

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2

Question

Decide which of the binary operations in the preceding exercise are commutative.

Proof

a. Isn't commutative since $a ⋆ b = a - b \neq b - a = b ⋆ a$ (ex: use $a = 1, b = 0$ )
b. Is commutative by the commutativity of addition and multiplication on $R$ :

\begin{aligned} a ⋆ b & = a + b + a b \\ = b + a + b a \\ = b ⋆ a \end{aligned}

c. This is commutative in a similar way to (b):

\begin{aligned} a ⋆ b & = \frac{a + b}{5} \\ = \frac{b + a}{5} \\ = b ⋆ a \end{aligned}

d. It's commutative using commutativity of addition and multiplication on $Z$ :

\begin{aligned} (a, b) ⋆ (c, d) & = (a d + b c, b d) \\ = (c b + d a, d b) \\ = (c, d) ⋆ (a, b) \end{aligned}

e. Clearly this is commutative since $1 ⋆ 2 = \frac{1}{2} \neq 2 = 2 ⋆ 1$ as a counter-example.

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6abcd

Question

Determine which of the following sets are Groups under addition:
a. The set of rational numbers (including 0 = 0/1) in lowest terms whose denominators are odd.
b. The set of rational numbers (including 0 = 0/1) in lowest terms whose denominators are even.
c. The set of rational numbers of absolute value < 1.
d. The set of rational numbers of absolute value $\geq 1$ together with 0.

Proof

a. This one is a group, since closure is occurring here. Namely, for any two fractions we add in this set it has the following format:

\frac{k_{1}}{2 m + 1} + \frac{k_{2}}{2 n + 1} = \frac{k_{1} (2 n + 1) + k_{2} (2 m + 1)}{4 m n + 2 (n + m) + 1}

where $k_{1}, k_{2}, m, n \in Z$ . Notice that the end fraction that comes out is odd on the bottom (since it equals $2 (2 m n + m + n) + 1$ ) and the top we don't care about (since it's definitely even, but the denominator is odd).

For the other properties:

Associativity comes from the associativity from $Q$
The identity is 0 since any element $q$ in this set follows $0 + q = q$ .
The inverse is just the negative of whatever rational integer, so $q$ has the inverse $- q$ .
Thus the set is a group.

b. This is not a group, because we fail to have closure under our binary operation. Namely, we may add two even-denominator fractions and get an odd one out. For example:

\frac{1}{2} + \frac{1}{2} = \frac{2}{2} \to \frac{1}{1}

is not in our set. Thus then this can't be a group.

c. Similar to (b) there isn't closure. For example:

\frac{1}{2} + \frac{1}{2} = 1

is not in the set since $1 ≮ 1$ .

d. This is the same as the last two:

\frac{3}{2} + \frac{- 3}{2} = 0

is not in the set while $\frac{3}{2}, \frac{- 3}{2}$ are.

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9

Question

Let $G = {a + b \sqrt{2} \in R | a, b \in Q}$ .
a. Prove $G$ is a Group under addition.
b. Prove that the nonzero elements of $G$ are a group under multiplication.

Proof

a. Clearly $G$ is closed under addition via an argument using closure of addition under $Q$ . For the other properties:

Associativity (let $x_{1}, x_{2}, x_{3} \in G$ ):

\begin{array}{r} (x_{1} + x_{2}) + x_{3} = x_{1} + (x_{2} + x_{3}) \end{array}

since $x_{1}, x_{2}, x_{3} \in R$ and are commutative under addition there.
2. The identity element is just $0 \in G$ . For any $x \in G$ we have $x + 0 = x$ .
3. The inverse of any $x = a + b \sqrt{2} \in G$ is just $- a - b \sqrt{2} \in G$ since:

\begin{aligned} x + (- a - b \sqrt{2}) & = (a + b \sqrt{2}) + (- a - b \sqrt{2}) \\ = (a - a) + (b - b) \sqrt{2} \\ = 0 + 0 \sqrt{2} \\ = 0 \end{aligned}

b. Clearly $G - {0}$ is closed under muliplication via how $Q - {0}$ is closed under multiplication. For the other properties:

Associativity. Same argument as the previous part. For $x_{1}, x_{2}, x_{3} \in G - {0}$ then $x_{1} x_{2} x_{3} = (x_{1} x_{2}) x_{3} = x_{1} (x_{2} x_{3})$ by associativity over $R$ .
The identity element is just $1 = 1 + 0 \sqrt{2} \in G - {0}$ . Notice that for any $x \in G - {0}$ that $x \cdot 1 = (a + b \sqrt{2}) \cdot (1 + 0 \sqrt{2}) = a + b \sqrt{2} = x$ .
The inverse of any element $x = a + b \sqrt{2} \in G - {0}$ depends on the values:
1. If $b = 0$ then $x = a$ so then $x^{- 1} = \frac{1}{a}$ since $x \cdot x^{- 1} = a \cdot \frac{1}{a} = 1$ as desired.
2. If $b \neq 0$ then $x^{- 1} = \frac{- a}{2 b^{2} - a^{2}} + \frac{b}{2 b^{2} - a^{2}} \sqrt{2}$ (notice that $2 b^{2} - a^{2} = 0 ⟺ \sqrt{2} b = a$ which is false since $\sqrt{2}$ is irrational. Verifying:

\begin{aligned} x \cdot x^{- 1} & = x^{- 1} \cdot x \\ = (\frac{- a}{2 b^{2} - a^{2}} + \frac{b}{2 b^{2} - a^{2}} \sqrt{2}) (a + b \sqrt{2}) \\ = \frac{- a^{2}}{2 b^{2} - a^{2}} + \frac{2 b^{2}}{2 b^{2} - a^{2}} + \sqrt{2} {(\frac{b a - a b}{2 b^{2} - a^{2}})}^{0} \\ = \frac{2 b^{2} - a^{2}}{2 b^{2} - a^{2}} \\ = 1 \end{aligned}

as desired.

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11

Question

Find the orders of each element of the additive group $\frac{Z}{12 Z}$

Proof

The identity element is $\overset{―}{0}$ . So for each element:

$| \overset{―}{0} | = 1$
$| \overset{―}{1} | = 12$
$| \overset{―}{2} | = 6$
$| \overset{―}{3} | = 4$
$| \overset{―}{4} | = 3$
$| \overset{―}{5} | = 12$
$| \overset{―}{6} | = 2$
$| \overset{―}{7} | = 12$
$| \overset{―}{8} | = 3$
$| \overset{―}{9} | = 4$
$| \overset{―}{10} | = 6$
$| \overset{―}{11} | = 12$
#todo/school : Try to learn about this type of group on pp. 10 (24) 📅 2025-01-08 ✅ 2025-01-08

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12

Question

Find the orders of the following elements of the multiplicative group ${(\frac{Z}{12 Z})}^{\times}$ : $\overset{―}{1}, \overset{―}{- 1}, \overset{―}{5}, \overset{―}{7}, \overset{―}{- 7}, \overset{―}{13}$ .

Proof

Again the identity is $\overset{―}{1}$ . Thus (here we use $\infty$ to denote an infinite order):

$| \overset{―}{1} | = 1$
$| \overset{―}{- 1} | = 2$
$| \overset{―}{5} | = 2$
$| \overset{―}{7} | = 2$
$| \overset{―}{- 7} | = 2$
$| \overset{―}{13} | = 2$

For the $2$ 's it's because the number $n^{2} \equiv 1 mod 12$ .

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25

Question

Prove that if $x^{2} = 1$ for all $x \in G$ then $G$ is abelian.

Proof

We know that $G$ is already a Group, so let's show that there's commutativity on $G$ . Let $a, b \in G$ . First notice that $x^{2} = x \cdot x = 1$ so then by definition then $x = x^{- 1}$ for all $x \in G$ . Thus:

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

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