prove that (Ha)(Hb)=Hab iff H is normal
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 $ghg^{-1}=h_0\in H$.
Ahb
Updated on December 22, 2020Comments
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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
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kccu almost 8 yearsHow is $(Ha)(Hb)$ being defined?
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Ahb almost 8 yearsthere is no more information in the question
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Jyrki Lahtonen almost 8 yearsWhy should the definition of $(Ha)(Hb)$ be in the question? Surely it is given in your textbook or lecture notes!
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Jyrki Lahtonen almost 8 yearsNot 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 :-)
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Ahb almost 8 yearsthanks 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
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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 $h' \in H$.