Understanding a proof about groups and cosets

  • Thread starter cragar
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  • #1
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Homework Statement


H is a subgroup of G, and a and b are elements of G.

Show that Ha=Hb iff [itex] ab^{-1} \in H [/itex] .

The Attempt at a Solution



line 1: Then a=1a=hb for some h in H.
then we multiply both sides by b inverse.
and we get [itex] ab^{-1}=h [/itex]
This is a proof in my book.
My question is on line 1 when they write 1a, is 1 the identity element in H.
So basically Ha=1a and this trick allows the rest to follow.
 

Answers and Replies

  • #2
CompuChip
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H is a subgroup, so the identity in G is also the identity in H.
 
  • #3
CompuChip
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Actually my answer was a bit short, so let me expand on it a bit more.

Recall the definition ##Ha \equiv \{ ha \mid h \in H \}##. Let's consider the '=>' direction first and assume that ##Ha = Hb##. Consider the fixed element ##a##. Since ##e \in H## (I prefer e over 1 to indicate the identity element), ##a = ea \in Ha##. On the other hand, since Ha = Hb, ##a \in Hb## meaning that there exists some ##h \in H## such that ##a = hb##. From this is follows that ##a b^{-1} (= h) \in H## which is all you needed to show.

I assume that after that they proceed to show the converse, i.e. if ##a b^{-1} \in H## for the given elements ##a, b## then ##Ha = Hb##. I imagine that proof will run along the lines of: assume ##ab^{-1} \in H##, let ##h \in H## be arbitrary, then ##h a = h a (b^{-1} b) = h (a b^{-1}) b## and since H is a subgroup, ##h (ab^{-1})## is the product of two elements in H which is itself in H, so ##h a = h' b## with ##h' ( = h a b^{-1}) \in H##, hence ##Ha \subseteq Hb## - the proof for ##Hb \subseteq Ha## follows similarly.

[edit]Note that actually ##\forall a, b \in G \left( a, b \in H \implies ab^{-1} \in H \right)## is a necessary and sufficient condition for H to be a subgroup of G. So basically your theorem is "Ha = Hb iff H is a subgroup" or "For subgroups, and subgroups only, all right cosets are equal".
 
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  • #4
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ok thanks
 

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