Proof involving group and subgroup

1. The problem statement, all variables and given/known data
If H is a subgroup of a group G, and a, b [itex]\in[/itex] G, then the following four conditions are equivalent:
i) [itex] ab^{-1} \in H [/itex]
ii) [itex] a=hb [/itex] for some [itex] h \in H [/itex]
iii) [itex] a \in Hb [/itex]
iv) [itex] Ha=Hb [/itex]


2. Relevant equations
cancellation law seems handy, and existence of an inverse, associativity, and other basic things.


3. The attempt at a solution

[itex] ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h \Rightarrow (ab^{-1})b=hb \Rightarrow a(bb^{-1})=ae=a=hb \Rightarrow a \in Hb \Rightarrow ...? [/itex]

 

does H=Hb?

 

I'm stuck. I tried doing something like a=hb but for ha=b but since h=ab^-1 that didn't help, so I considered [itex] h^{-1}=ba^{-1} [/itex] so [itex] ba^{-1}=h^{-1} [/itex] and [itex] h=ab^{-1} \Rightarrow [/itex]
But I'm not getting anywhere. What should come directly after [itex] a \in Hb \Rightarrow [/itex] ?

 

is it a problem that I'm looking for Ha, not aH?

 

okay, h is in H already, so hh is in H, and [itex] hh=hab^{-1}=h' \in H [/itex]

 

so [itex] h' \in H \Rightarrow h'b \in Hb \Rightarrow ah'b \in Hb \Rightarrow ah' \in H \Rightarrow [/itex]
[itex] h^{-1}ah' \in H \Rightarrow ba^{-1}ah' \in H \Rightarrow bh' \in H [/itex]

 

Because [itex] ab^{-1}=h [/itex] so [itex] hh=ab^{-1}h [/itex]

 

the very beginning, [itex] ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h [/itex]

 

okay I'll pick a different h but I disagree about calling it a "previous proof" since the whole point of this exercise is to make the thing a connected circle.
so thats why I need something to follow directly from a \in Hb.

but first it's good to break it down into just 3 => 4
so, [itex] h \in H \Rightarrow ha \in Ha [/itex]
Now, choosing [itex] h' : h'=hab^{-1} [/itex] is in H because we assumed [itex] ab^-1 \in H [/itex]
[itex] ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow h' \in Hh [/itex] crap I seem lost again.

 

I'm breaking out the chocolate fudge brownie ice cream now. Are you trying to make me cry? I have a lot of other proofs to study before test Tuesday :(

 

what? Where?
[itex] ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow hab^{-1} \in H [/itex] but how does that show [itex] ha=hb^{-1} [/itex] ?