subgroup

Subgroup

A subgroup is a substructure which also has group properties:
$(G,\circ )$ group, $U\subseteq G$ subgroup $⟺(U,\circ )$ group.
We denote it as $U\le G$.

EXAMPLE

$S_{\square ,\circ }\le S_4$
$(\mathbb{Q}\setminus \{0\})\le (\mathbb{R}\setminus \{0\},\cdot )$
$(m\mathbb{Z},+)\le (\mathbb{Z},+)$
$(\mathbb{N},+)\nleqq (\mathbb{Z},+)$

Criteria for being a subgroup

Let $(G,\circ )$ group, $U\subseteq G⟹U\le G⟺$
(1) $U\ne \varnothing$
(2) $\forall x,y\in U:x\circ y\in U$
(3) $\forall x\in U:x^{-1}\in U$
$⟺$
(1) $U\ne \varnothing$
(2) $\forall x,y\in U:x\circ y^{-1}\in U$ (for finite groups, just regular closure is enough)
We only need to check these two, to know that $U$ is also a group.

Proof: $U\le G⟺U\ne \varnothing \wedge (\forall x,y\in U:x\circ y^{-1}\in U$
“$\Rightarrow$”: $e\in U⟹U\ne \varnothing$, $x,y\in U⟹x\circ y^{-1}\in U$ (because $G$ is a group, and $U$ lies within it, it still has these properties)
“$\Leftarrow$”:
Law of associativity is also inherited from the parent group, $\exists x\in U⟹x\circ x^{-1}\in U⟹e\in U⟹$ neutral element exists
$x=e,y=a,e,a\in U⟹e\circ a^{-1}=a^{-1}\in U⟹$ inverse exists (because $U$ is part of $G$ which is a group), and we just showed that it is also in the subgroup $U$, meaning we can invert within the subgroup.
$x,y\in U⟹x,y^{-1}\in U⟹x\circ (y^{-1}{)}^{-1}\in U$ (closure)

Prove $(m\mathbb{Z},+)\le (\mathbb{Z},+)$

${\mathbb{Z}}_m=\{\bar{0},\bar{1},\dots \overline{m-1}\}$
(1) $0\in m\mathbb{Z}⟹m\mathbb{Z}\ne \varnothing$
(2) $a,v\in m\mathbb{Z}⟹\exists k,l\in \mathbb{Z}:a=km,b=lm$
$a+(-b)\in m\mathbb{Z}:a-b=km-lm=(k-l)m\in \mathbb{Z}⟹m\mathbb{Z}\le \mathbb{Z}$

A coset is what you get when you take every element of a subgroup and multiply it (from the left or right) by a fixed group element. It represents a kind of "shifted copy" of the subgroup within the larger group.

The direction you multiply from (left or right) can give different results if the group is not commutative.

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The set of all powers of $a$: $⟨a⟩:=\{a^n\mid n\in \mathbb{Z}\}$

Let $(G,\circ )$ be a group and $\exists a\in G:⟨a⟩=G$, then $⟨a⟩$, is a cyclic subgroup of $G$, generated by $a$ (“von $a$ erzeugte Untergruppe”).

$a\le G$, … is it actually a subgroup?
Note empty: $a^0=e\in ⟨a⟩⟹⟨a⟩\ne \varnothing$
Closed: $x,y\in ⟨a⟩⟹\exists n,m\in \mathbb{Z}:x=a^n,y=a^m⟹x\circ y=a^n{a}^m=a^{n+m}\in ⟨a⟩$ → $⟨a⟩$ abelian group
Invertible: $x\in ⟨a⟩⟹\exists n\in \mathbb{Z}:x=a^n⟹x^{-1}=a^{-n}⟹a^{-n}\in ⟨a⟩$ (because $a^n{a}^{-n}=a^0=e)$
$✓$

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