Subgroups and Cosets
Definition 1.7 – Subgroup
A subset $H$, of a group $G$, sharing the same product, is a subgroup of $G$.
Each group has at least one subgroup: the group itself and the identity. These are called improper subgroups. Any other subgroup is called a proper subgroup.
Definition 1.8 – Cosets
Consider $S = \{ s_{1}, s_{2}, \dots, s_{k}\}$, a finite subset of $G$, and $g \in G$. We define
- $Sg = \{ s_{1}g, s_{2}g, \dots, s_{k}g \}$, the right coset
- $gS = \{ gs_{1}, gs_{2}, \dots, gs_{k}\}$, the left coset
If there is no distinction between $Sg$ and $gS$, we simply refer to both as cosets.
Note that the operations $Sg$ and $gS$ also apply to subsets, in which case they are not cosets.
Theorem 1.1
A finite subgroup of a finite or infinite, group, which is respects closure under multiplication, is a subgroup.
Theorem 1.2
Consider some finite subgroup (or subset) $S$ of $G$, with $g \in G$. The number of distinct elements in $Sg$, or $gS$, is equal to the number of distinct elements in $S$.
Proof
Suppose the theorem is true. Then it must mean that $s_{a}g = s_{b}g$ if, and only if, $s_{a} = s_{b}$. Otherwise, $Sg$ would have duplicate elements and the order $\lvert Sg \rvert$ would be less than $\lvert S \rvert$.
- If $s_{a} = s_{b}$, then $s_{a}g = s_{b}g$.
- If $s_{a}g = s_{b}g$, then $s_{a}gg^{-1} = s_{b} gg^{-1} \implies s_{a} = s_{b}$.
The same can be shown for the left coset.
Theorem 1.3 – Rearrangement Theorem
Consider some, finite or infinite, group $G$.
$$ \forall g \in G, \quad Gg = gG = G $$This means that, in the multiplication table of $G$, each element appears once, and only once, in each row and each column.
Proof
The proof is not dissimilar to that of theorem 1.2; we can show that, $\forall g \in G$, $g_{1}g = g_{2}g$ if, and only if, $g_{1} = g_{2}$. Otherwise, the order of $Gg$ or $gG$ would be less than $\lvert G \rvert$, meaning the groups are not equal.
Theorem 1.4 – Lagrange’s Theorem
The order of a subgroup $H$, of a finite group $G$, divides the order of $G$. In other words, $\lvert G \rvert$ is a multiple of $\lvert H \rvert$.
Corollary : if a group has an odd order, it necessarily has no subgroup.
Proof
Consider the group $G = \{ g_{1}, g_{2}, g_{3}, \dots, g_{n} \}$, of order $\lvert G \rvert = n$, and one of its subgroups $H = \{h_{1}, h_{2}, h_{3}, \dots, h_{m}\}$, of order $\lvert H \rvert= m$, with $g_{1} = h_{1} = e$.
First Statement
The set of all right coset $Hg$, for every $g \in G$, contains $n$ cosets. Each element $g_{a} \in G$ is in at least one of the cosets
This is because $e \in H$.
Second Statement
If we select two cosets, at random, they are either identical or disjoint.
Suppose $Hg_{a}$ and $Hg_{b}$ ($g_{a}, g_{b} \in G$) have at least one element in common; $h_{c}g_{a} = h_{d}g_{b}$ for some $h_{c},h_{d} \in H$. A single coset would then be formed in comparison to this element.
$$ H(h_{c}g_{a}) = H(h_{d}g_{b}) $$By associativity,
$$ (Hh_{c})g_{a} = (Hh_{d})g_{b} $$The rearrangement theorem tells us that $Hh_{c} = Hh_{d}$, so,
$$ Hg_{a} = Hg_{b} $$Thus, $Hg_{a}$ is identical to $Hg_{b}$, meaning the two cosets are either identical or disjoint.
Proof (cont.)
There are $n$ cosets of $H$, each with $m$ elements; some are identical and others are disjoint. If we eliminate the redundant ones, we are left with $p$ disjoint cosets.
Because of the first statement, together, they contain all the elements of $G$ once, and only once. So, the $pm$ elements of these distinct cosets are the same $n$ elements of $G$.
$$ \therefore \quad m = \frac{n}{p} \implies \lvert G \rvert = p \lvert H \rvert $$Corollary
Since the order of some group $G$ with some subgroup $H$ is a multiple of $\lvert H \rvert$, any group of odd order has no subgroup.