Is it a subgroup of G?

Algebra Level pending

( G , ⋅ ) (G, \cdot ) group, H ≤ G , g ∈ G H \leq G, g \in G .

Could we say that g − 1 H g : = { g − 1 h g ; h ∈ H } g^{-1}Hg := \{ g^{-1}hg ; h \in H \} is subgroup of G G ? that is g − 1 H g ≤ G g^{-1}Hg \leq G ?

True False

Icaro Buscarini by Icaro Buscarini , 26, Brazil

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Icaro Buscarini
Sep 7, 2018

Answer: True

proof:

∙ e G ∈ g − 1 H g ‾ \bullet \hspace{2mm} \underline{e_G \in g^{-1}Hg} :

H ≤ G ⇒ e G ∈ H \hspace{4mm} H \leq G \Rightarrow e_G \in H

g − 1 e G g = g − 1 g = e G \hspace{4mm} g^{-1}e_{G}g = g^{-1}g = e_{G}

∴ e G ∈ g − 1 H g \hspace{4mm} \therefore e_{G} \in g^{-1}Hg

∙ a , b ∈ g − 1 H g ⇒ a b ∈ g − 1 H g ‾ \bullet \hspace{2mm} \underline{a, b \in g^{-1}Hg \Rightarrow ab \in g^{-1}Hg} :

a ∈ g − 1 H g ⇒ ∃ h 1 ∈ H \hspace{4mm} a \in g^{-1}Hg \Rightarrow \exists h_1 \in H , such that a = g − 1 h 1 g a = g^{-1}h_1g

b ∈ g − 1 H g ⇒ ∃ h 2 ∈ H \hspace{4mm} b \in g^{-1}Hg \Rightarrow \exists h_2 \in H , such that b = g − 1 h 2 g b = g^{-1}h_2g

\hspace{4mm} Then a b = ( g − 1 h 1 g ) ( g − 1 h 2 g ) = g − 1 h 1 ( g g − 1 ) h 2 g = g − 1 h 1 ( e G ) h 2 g = g − 1 h 1 h 2 g ab = (g^{-1}h_1g)(g^{-1}h_2g) = g^{-1}h_1(gg^{-1})h_2g = g^{-1}h_1(e_{G})h_2g = g^{-1}h_1h_2g

H ≤ G ⇒ h 1 h 2 ∈ H ⇒ g − 1 h 1 h 2 g ∈ g − 1 H g \hspace{4mm} H \leq G \Rightarrow h_1h_2 \in H \Rightarrow g^{-1}h_1h_2g \in g^{-1}Hg

∴ a b ∈ g − 1 H g \hspace{4mm} \therefore ab \in g^{-1}Hg

∙ ∀ a ∈ g − 1 H g , ∃ b ∈ g − 1 H g ⇒ a b = e G ‾ \bullet \hspace{2mm} \underline{\forall a \in g^{-1}Hg, \exists b \in g^{-1}Hg \Rightarrow ab = e_{G}} :

a ∈ g − 1 H g ⇒ ∃ h ∈ H \hspace{4mm} a \in g^{-1}Hg \Rightarrow \exists h \in H , such that a = g − 1 h g a = g^{-1}hg

H ≤ G ⇒ h − 1 ∈ H \hspace{4mm} H \leq G \Rightarrow h^{-1} \in H , take b = g − 1 h − 1 g , b ∈ g − 1 H g b = g^{-1}h^{-1}g, b \in g^{-1}Hg

\hspace{4mm} Then a b = ( g − 1 h g ) ( g − 1 h − 1 g ) = g − 1 h ( g g − 1 ) h − 1 g = g − 1 h ( e G ) h − 1 g = g − 1 h h − 1 g = g − 1 e G g = g − 1 g = e G ab = (g^{-1}hg)(g^{-1}h^{-1}g) = g^{-1}h(gg^{-1})h^{-1}g = g^{-1}h(e_{G})h^{-1}g = g^{-1}hh^{-1}g = g^{-1}e_{G}g = g^{-1}g = e_{G}

Conclusion : g − 1 H g ≤ G g^{-1}Hg \leq G

2 Helpful 1 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...