Left (and Right) Cosets form a Subgroup

Theorem

Let $G$ be a Group and let $K$ be the Kernel of some homomorphism from $G$ to another group. Then the set whose elements are the left cosets of $K$ in $G$ with operation defined by:

u K \circ v K = (u v) K

forms a group $\frac{G}{K}$ , a Quotient Group. In particular, this operation is well defined where any $u_{1} \in u K$ and $v_{1} \in v K$ then $u_{1} v_{1} \in u v K$ so then $u_{1} v_{1} K = u v K$ (so multiplication doesn't depend on the choice of representatives for the cosets). The same is said using "right coset" instead of "left coset".

Proof

Let $X, Y \in \frac{G}{K}$ and let $Z = X Y$ in $\frac{G}{K}$ , so that by Properties of Quotient Groups then $X, Y, Z$ are (left) cosets of $K$ . By assumption, $K$ is the kernel of some Homomorphism (or Group Morphism) $φ : G \to H$ so then $X = φ^{- 1} (a)$ , $Y = φ^{- 1} (b)$ for some $a, b \in H$ .

By definition of the operation in $\frac{G}{K}$ then $Z = φ^{- 1} (a b)$ . Letting $u, v$ be arbitrary representatives of $X$ and $Y$ respectively, we get that $φ (u) = a$ and $φ (v) = b$ and $X = u K, Y = v K$ . We must show that $u v \in Z$ to show well-definedness. Note:

\begin{aligned} u v \in Z & ⟺ u v \in φ^{- 1} (a b) \\ ⟺ φ (u v) = a b \\ ⟺ φ (u) φ (v) = a b \end{aligned}

which holds since $φ (u) = a$ and $φ (v) = b$ . Thus then $u v \in Z$ .

A later exercise shows that every $z \in Z$ has some $u \in X$ and $v \in Y$ where $z = u v$ .

This all proves that the product of $X$ with $Y$ is the coset $u v K$ for any choice of representatives $u, v$ , showing well-definedness.

To show it works for "right" cosets, use Properties of Quotient Groups to show that $u K = K u$ and $v K = K u$ for all $u, v \in G$ , so then $K u \circ K v = K (u v)$ using the "left" coset part we just proved.

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