Question

Let G be a group, H⍙G and β ≤ G/H . Let A = {x ∈G | Hx∈β} . Show that i) A ≤ G , ii) H⍙A , iii) β = A/H.

Answer 1

Answer on Question #82407 – Math – Abstract Algebra

Question

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

i) $A \leq G$

ii) $H \triangle A$

iii) $\beta = A / H$

Solution

Let $f$ be homomorphism $G \to G / H$.

i) $A = \{x, f(x) \in \beta\}$ so $A = f^{-1}(\beta)$, so $A$ is a subgroup of $G$, because a preimage of a subgroup is a subgroup.

ii) $H = f^{-1}(0)$, so $H \leq A$, and $H$ is normal in $G$, so $\forall h \in H$ and $\forall g \in G \Rightarrow ghg^{-1} \in H$, so even more so $\forall h \in H$ and $\forall a \in A \Rightarrow aha^{-1} \in H$, so $H$ is normal in $A$: $H \triangleleft A$.

iii) Let $Ha \in A / H$, so $a \in A$, so $Ha \in \beta$ (by definition), so $A / H \subseteq \beta$.

And vice versa, $Hb \in \beta \Rightarrow b \in A \Rightarrow Hb \in A / H \Rightarrow \beta \subseteq A / H$

So $A / H \subseteq \beta$ and $\beta \subseteq A / H \Rightarrow \beta = A / H$.

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