📂Abstract Algebra

Quotient groups in Abstract Algebra

Definition ¹

Let’s call the set of all cosets of $H \subset G$ as $G / H$. If there exists a well-defined binary operation $\ast$ like $(aH) \ast\ (bH) = (ab) H$, then $\left< G / H , * \right>$ is called a Factor group.

Theorem

Let’s assume $H \leqslant G$. That $H \triangleleft G$ and $G / H$ being a group are equivalent.

Description

That $H \triangleleft G$ means that $H$ is a normal subgroup of $G$.

The binary operation $\ast$ is a binary operation that calculates only with the representative elements of cosets, which makes the set $G / H$ form a factor group. If it’s not intuitively understandable why $G / H$ forms a group, it’s highly likely that the concept of cosets is misunderstood.

Proof

It suffices to show that $( \implies )$ $(aH) (bH) = (ab) H$.

Since $H$ is a normal subgroup, if $h_{1} b \in H b$, then there exists some $h_{3} \in H$ such that $b h_{3} \in bH$. For $ah_{1} \in aH$ and $bh_{2} \in H$ $$(ah_{1}) (b h_{2}) = a(h_{1} b)h_{2} = a b h_{3} h_{2} = ab (h_{3} h_{2}) \in (ab) H$$ Therefore, $(aH) (bH) \subset (ab) H$, and reversing the process, $(ab) H \subset (aH) (bH)$ thus $$(aH) (bH) = (ab) H$$

---

It suffices to show that $( \impliedby )$ $gH = Hg$.

If we have $x \in gH$ and $g^{-1} \in g^{-1} H$, then $$(xH) (g^{-1} H) = (x g^{-1}) H$$ thus, $h := x g^{-1} \in H$ must be the case. Meanwhile, since $x = hg$, $$x \in Hg$$ Therefore, $gH \subset Hg$, and reversing the process, $Hg \subset gH$ thus $$gH = Hg$$

■