prove that (Ha)(Hb)=Hab iff H is normal

1,856

The set $(Ha)(Hb)$ consists of all elements of the form $h_1ah_2b$, for $h_1,h_2\in H$.

Suppose $H$ is normal in $G$. We want to show that $(Ha)(Hb)=Hab$.

Proof that $(Ha)(Hb)\subseteq Hab$. Let $h_1,h_2\in H$. Then $h_1ah_2b=h_1ah_2a^{-1}ab$. If $h=h_1ah_2b=h_1ah_2a^{-1}=h_1(ah_2a^{-1})$ we see that, by normality, $h\in H$. Therefore $h_1ah_2b=hab\in Hab$.

Proof that $Hab\subseteq (Ha)(Hb)$. Let $h\in H$; then $hab=1a(a^{-1}ha)b$. Set $h_1=1\in H$ and $h_2=a^{-1}ha\in H$; then $hab=h_1ah_2b\in(Ha)(Hb)$

Suppose $(Ha)(Hb)=Hab$, for all $a,b\in G$. We want to show that $H$ is normal in $G$.

Let $g\in G$ and $h\in H$. Then $ghg^{-1}=1ghg^{-1}\in (Hg)(Hg^{-1})$. Since $(Hg)(Hg^{-1})=Hgg^{-1}$ we know there exists $h_0\in H$ such that $$ ghg^{-1}=h_0gg^{-1} $$ Therefore $ghg^{-1}=h_0\in H$.

Share:
1,856
Ahb
Author by

Ahb

Updated on December 22, 2020

Comments

  • Ahb
    Ahb almost 3 years

    if H is a subgroup of G prove that (Ha)(Hb)=Hab for all right cosets Ha and Hb of H in G iff H is normal in G

    • kccu
      kccu almost 8 years
      How is $(Ha)(Hb)$ being defined?
    • Ahb
      Ahb almost 8 years
      there is no more information in the question
    • Jyrki Lahtonen
      Jyrki Lahtonen almost 8 years
      Why should the definition of $(Ha)(Hb)$ be in the question? Surely it is given in your textbook or lecture notes!
    • Jyrki Lahtonen
      Jyrki Lahtonen almost 8 years
      Not sure whether it is best to close this as being off-topic for lack of context or as a dup. I was searching for a suitable duplicate. Found that "gem", and stopped looking. If you feel another closure reason is better, I can fix that :-)
  • Ahb
    Ahb almost 8 years
    thanks alot but what i did not anderstand is "for any h∈Hh∈H, there exists h′∈Hh′∈H such that ghg−1=h′ghg−1=h′" why is that
  • angryavian
    angryavian almost 8 years
    @Ahb It follows from the equality $H g^{-1} H g = H$. $1g^{-1} h g$ is an element on the left-hand side, and the equality implies that it is equal to some $h' \in H$.