Question

Let G be a group , H ∆= G and beta <=(G /H ) . Let A = { x belongs to G | Hx belongs to beta } . Show that (1) . A<=G (2).H ∆= A (3). Beta = (A\H ) .

Answer 1

Answer on Question #78753 - Math - Abstract Algebra

June 30, 2018

Question. Let $G$ be a group, $H \triangleleft G$ and $\beta \leq (G / H)$. Let $A = \{x \in G \mid Hx \in \beta\}$. Show that

1. $A \leq G$,

2. $H \triangleleft A$,

3. $\beta = (A / H)$.

Answer.

1. We will show that $A$ is closed under group operations of $G$.

- Let $x, y \in A$. Then $Hx, Hy \in \beta$ by the definition of $A$. As $\beta \leq (G / H)$, $HxHy \in \beta$. As $H \triangleleft G$, $HxHy = HHxy = Hxy$. Hence $xy \in A$ by the definition of $A$.

- As $He \in \beta$, $e \in A$.

- Let $x \in A$. Then $Hx \in \beta$ by the definition of $A$. As $\beta \leq (G / H)$, $x^{-1}H = x^{-1}H^{-1} = (Hx)^{-1} \in \beta$. As $H \triangleleft G$, $x^{-1}H = Hx^{-1}$. Hence $x^{-1} \in A$ by the definition of $A$.

2. Let $x \in A$. Then $x \in G$, and $xH = Hx$ because $H$ is normal in $G$. Therefore, $H$ is normal in $A$.

3. We know that $(A / H) = \{Hx \mid x \in A\}$. We will prove $\beta = (A / H)$ which is equivalent to

$$x' \in \beta \iff x' \in (A / H)$$

for every $x' \in (G / H)$.

- ( $\Longrightarrow$ ) Let $x' \in \beta$. Then $x' = Hx$ for some $x \in G$ because $\beta \leq (G / H)$. Then $x \in A$ by the definition of $A$. Therefore, $x' \in (A / H)$ by the definition of $(A / H)$.

- ( $\Longleftarrow$ ) Let $x' \in (A / H)$. Then $x' = Hx$ for some $x \in A$ by the definition of $(A / H)$. Therefore, $x' \in \beta$ by the definition of $A$.

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