📂Abstract Algebra

Factor Group in Abstract Algebra

Definition¹

Let $G / H$ be the set of all cosets of $H \subset G$. If there exists a well-defined binary operation $\ast$ such that $(aH) \ast\ (bH) = (ab) H$, then $\left< G / H , * \right>$ is called a factor group.

Theorem

Suppose $H \leqslant G$. Then $H \triangleleft G$ is equivalent to $G / H$ being a group.

Explanation

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

The binary operation $\ast$ is a binary operation computed only on the representatives of the cosets, and it makes the set $G / H$ form a factor group. If it is not intuitively clear why $G / H$ becomes a group, there is a good chance that the concept of cosets has been misunderstood from the start.

Proof

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

Since $H$ is a normal subgroup, if $h_{1} b \in H b$ then there exists some $h_{3} \in H$ satisfying $b h_{3} \in bH$. 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 above process gives $(ab) H \subset (aH) (bH)$, so $$ (aH) (bH) = (ab) H $$

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

Let $x \in gH$ and $g^{-1} \in g^{-1} H$. Then $$ (xH) (g^{-1} H) = (x g^{-1}) H $$ so it must be that $h := x g^{-1} \in H$. Meanwhile, since $x = hg$, $$ x \in Hg $$ Therefore $gH \subset Hg$, and reversing the above process gives $Hg \subset gH$, so $$ gH = Hg $$

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