Properties of Quotient Groups

Theorem

Let $φ : G \to H$ be a Homomorphism (or Group Morphism) of Groups with Kernel $K$ . Let $X \in \frac{G}{K}$ be the Fiber (Group Theory) above $a$ , so then $X = φ^{- 1} (a)$ . Then:

Proof

(1): Let $u \in X$ , so then by the definition of $X$ then $φ (u) = a$ . Let:

u K = {u k | k \in K}

So all we have to show is that $u K = X$ :

\begin{aligned} φ (u k) & = φ (u) φ (k) \\ = a φ (k) \\ = a \cdot 1 & (k \in K = Ker (φ) = {g \in G | φ (g) = 1_{H}}) \\ = a \end{aligned}

\begin{aligned} φ (k) & = φ (u^{- 1}) φ (g) \\ = φ (u)^{- 1} φ (g) \\ = a^{- 1} a \\ = 1_{H} \end{aligned}

so then $k \in Ker (φ) = K$ . Thus then $g = u k \in u K$ as desired.

(2): Left as an exercise to the reader.

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These sets often come up, so we defined them as Left and Right Cosets.