Last time we have introduced laws of composition $(S, )$ and defined what it means for $$ to be associative/commutative and to admit an identity.
Compare the following proposition to problem #5 from the first homework.
Let $(S, *)$ be a set with an associative composition law and identity. Let $x, y \in S$ be two invertible elements, then $x * y$ is also invertible with $(x * y)^{-1}=y^{-1} * x^{-1}$
Proof. Indeed, we have
$$
(x * y) *\left(y^{-1} * x^{-1}\right)=x *\left(y * y^{-1}\right) * x^{-1}=x * e * x^{-1}=x * x^{-1}=e
$$
where in the first equality we used associativity, in the second the fact that $y^{-1}$ is the inverse of $y$, in the third the property of the identity $e$, and in the last the fact that $x^{-1}$ is the inverse of $x .$
Similarly we check that $\left(y^{-1} * x^{-1}\right) *(x * y)=e$ which proves that $\left(y^{-1} * x^{-1}\right)=(x * y)^{-1}$.
Groups is one of the most fundamental notions in mathematics. Whenever you have an object (of any nature) admitting symmetries, there is a group lurking behind.
4. Example 1: Key motivating example
Let $\Delta$ be an equilateral triangle. Then there 6 rigid motions mapping $\Delta$ onto itself, i.e., bijections $f: \Delta \rightarrow \Delta$ preserving the distances between points:
- 3 rotations – by $0^{\circ}, 120^{\circ}$ and $240^{\circ}$ (denote them $\mu_{0}, \mu_{1}, \mu_{2}$ )
- 3 reflections along 3 altitudes $l_{1}, l_{2}, l_{3}$ (denote the reflection by $\sigma_{1}, \sigma_{2}, \sigma_{3}$ ).
We claim that the composition operation on the set $\left{\mu_{0}, \mu_{1}, \mu_{2}, \sigma_{1}, \sigma_{2}, \sigma_{3}\right}$ defines a law of composition (exercise).
Problem 1: Prove that the above law of composition is associative, has an identity and every element admits an inverse. Show that it is not commutative.
Let $(G, *)$ be a set with a law of composition. We will say that $(G, *)$ is a group if the following properties hold:
$(\mathrm{G} 1) *$ is associative
(G2) there is an identity $e \in G$
(G3) every element $x \in G$ is invertible.
A group $(G, )$ is commutative or abelian if $$ is commutative.
7. Proposition 2: Cancellation law
If $(G, *)$ is a group and $x, y, z \in G$ are elements such that
$$
x * y=x * z
$$
then $y=z$, i.e., we can cancel out $x$ from the both sides.
Proof. Let us multiply both sides of the above identity by $x^{-1}$ from the left. Then we will have
$$
x^{-1} * x * y=x^{-1} * x * z .
$$
Due to associativity we do not have to specify the brackets and can perform multiplication in any order we like (without swapping the elements). Then on both sides we have $x^{-1} * x=e$, so
$$
e * y=e * z
$$
Now, due to the definition of $e$, we have $y=z$.
For $n \in \mathbb{Z}$ we will write
$$
x^{n}:=\underbrace{x * x * \cdots * x}_{n \text { times }}
$$
if $n>0$ and
$$
x^{n}:=(\underbrace{x * x * \cdots * x}_{-n \text { times }})^{-1}
$$
if $n<0$. As usual, $x^{0}=e$.
Check that for $n, m \in \mathbb{Z}$ we have $\left(x^{n}\right)^{m}=\left(x^{m}\right)^{n}=x^{n m}$
To feel better the notion of a group we will need to stock on examples.
$(\mathbb{Z},+)$, integers under addition form a group with the neutral element being 0 .
$(\mathbb{Z}, \times)$ is not a group, since 2 does not have a multiplicative inverse in $(\mathbb{Z}, \times) .$
$(\mathbb{R}, \times)$ is not a group, since 0 does not have a multiplicative inverse.
$(\mathbb{R} \backslash{0}, \times)$ is a group with neutral element being 1 .
$\left(\mathbb{Z}_{n},+\right)$ is a group with neutral element $[0]$.
$\left(\mathbb{Z}{n}, \times\right)$ is not a group since $[0]$ does not have a multiplicative inverse, however, similarly to the example of $(\mathbb{R} \backslash{0}, \times)$, we find that the set of units $\left(\mathbb{Z}{n}^{\times}, \times\right)$is a group.
All of the above groups are commutative.