4.1 Groups

Every definition of an abstract object consists of two parts:

1. in the first part we list all the data required for the definition,

2. in the second part we list all the axioms the data must satisfy.

Groups

Data:

• A set $G$

• An operation $○:G\times G\to G$

Axioms:

• The operation $○$ is associative.

• The operation $○$ has a neutral element.

• Every element $x\in G$ has an inverse.

In this case, we say that the pair $(G,○)$ is a group. In order to simplify the notation, we usually say simply that $G$ is a group assuming that the operation in use is clear. If in addition we have

• The operation $○$ is commutative.

Then the group $G$ is called abelian or simply commutative.

In short, a group is a set with a good operation.

Here, good means that we do not care about parentheses, we have neutral element and every element is invertible but the order of the elements still matters.

Abelian group means that additionally the order of the elements does not matter.

Subgroups

Let $G$ be a group. We define a subgroup $H$ of $G$.

Data:

• A subset $H\subseteq G$.

Axioms:

• The neutral element $1$ of $G$ belongs to $H$.

• $xy\in H$whenever$xy\in H$.

• $x^{-1}\in H$whenever$x\in H$.

In this case, we say that $H$ is a subgroup of $G$.

It should be noted that if $H$ is a subgroup of $G$, then $\cdot$ is a well-defined operation on $H$ and $(H,\cdot )$ becomes a group.

Cosets

Algebra usually tends to study groups using subgroups rather than elements. The main tool here is cosets.

Let $G$ be a group, $H\subseteq G$ a subgroup and $g\in G$ an arbitrary element. Then the set

$gH=\{gh|h\in H\}$

is called the left coset of $H$ with respect to $g$. In a similar way, we define right cosets. The set

$Hg=\{hg|h\in H\}$

is called the right coset of $H$ with respect to $g$.

The Lagrange Theorem

Let $G$ be a finite group and $H\subseteq G$ be a subgroup. Then, $\left|G\right|=(G:H)\left|H\right|$

Corollaries of The Lagrange Theorem

1. Let $G$ be a finite group and $H\subseteq G$ be a subgroup. Then $\left|H\right|$ divides $\left|G\right|$.

2. Let $G$ be a finite group and $g\in G$ be an arbitrary element. Then ord$\left(g\right)$ divides $\left|G\right|$. Indeed, ord$\left(g\right)=\left|⟨g⟩\right|$. But $\left|⟨g⟩\right|$ divides $\left|G\right|$ by the previous item.

3. Let $G$ be a finite group and $g\in G$ be an arbitrary element. Then $g^{\left|G\right|}=1$. Indeed, we already know that $\left|G\right|=$ord$\left(g\right)k$. Hence,

   $g^{\left|G\right|}=g^{\mathrm{ord}\left(g\right)k}={\left(g^{\mathrm{ord}\left(g\right)}\right)}^k=1^k=1$

4. Let $G$ be a group of prime order $p$. Then, $G$ is cyclic. Indeed, since the order of $G$ is prime, it is greater than $1$. Hence, there is an element $g\in G$ such that $g\ne 1$. Hence $⟨g⟩$ has order greater than $1$. But $\left|⟨g⟩\right|$ divides $\left|G\right|=p$. Since $p$ is prime, the only option is $\left|⟨g⟩\right|=p=\left|G\right|$. The latter means that $⟨g⟩=G$ and we are done.

5. The Fermat Little Theorem. Let $p\in \mathbb{Z}$ be a prime number and $a\in \mathbb{Z}$. If $p$ does not divide $a$, then $p$ divides $a^{p-1}-1$. Indeed, let us consider the group $({{\mathbb{Z}}_p}^{\ast },\cdot )$. For any element $b\in {{\mathbb{Z}}_p}^{\ast }$, we have $b^{\left|{{\mathbb{Z}}_p}^{\ast }\right|}=1\mathrm{mod}\left(p\right)$ by item~$3.$ But ${{\mathbb{Z}}_p}^{\ast }$ has $p-1$ elements. Now, let $a\in \mathbb{Z}$ be comprime with $p$. We denote its remainder modulo $p$ by $b$. Then $a^{p-1}=b^{p-1}=1\mathrm{mod}\left(p\right)$ and we are done.