Can't quite see why $b^{-1}a \in H \implies Ha^{-1}=Hb^{-1}$ I'm trying to prepare for my first "real" math class next semester in abstract algebra and I am stuck trying to prove the following implications
$$
aH=bH \implies a^{-1}b \in H \implies a^{-1}(b^{-1})^{-1} \in H \implies Ha^{-1}=Hb^{-1}
$$
and I'm pretty sure the details needed to verify the first two implications are 
$$
aH=bH \implies a^{-1}(aH)=a^{-1}bH \implies H=a^{-1}bH \implies a^{-1}b \in H 
$$
with the last one resulting from the fact that $e \in H$ (since $H$ is a subgroup) so by letting $h=e$ we get  $a^{-1}be \in H$ but for the second implication, I am less sure of the reasoning
$$
a^{-1}b \in H \implies (a^{-1}(b^{-1})^{-1})^{-1} \in H \implies b^{-1}a \in H \implies Hb^{-1}a=H \implies Hb^{-1}=Ha^{-1}
$$
So basically, I am wondering why it is possible to reverse the implication from 
$$
aH=bH \implies a^{-1}b \in H 
$$
to 
$$
b^{-1}a \in H \implies Hb^{-1}=Ha^{-1}
$$
The only way I could see this is if you write 
$$
(a^{-1}(b^{-1})^{-1}H)^{-1} \subseteq H \implies Hb^{-1}a \subseteq H \implies Hb^{-1} \subseteq Ha^{-1}
$$
but then I get stuck trying to prove that $Ha^{-1} \subseteq Hb^{-1}$. So is it permissable to "cross" the set membership relation $\in$ in the same way you would with an "=" so that something like 
$$
b^{-1}a \in H \implies b^{-1} \in Ha^{-1}
$$
and then maybe multiply by an arbitrary $h \in H$  so that $hb^{-1} \in hHa^{-1}$ and then since $HH=H$ we'd have $Hb^{-1}=Ha^{-1}$? Maybe I'm over thinking it or I missed something obvious so any hints would be very helpful. I just don't want to gloss over something that I don't fully understand.
 A: You want to show that $aH=bH \implies Ha^{-1}=Hb^{-1}$. 
Some preliminaries first:
By definition of cosets $aH=\{ah, \,| \, h \in H \}$, likewise $bH=\{bh, \,| \, h \in H \}$
The set equality $aH=bH$, implies $ah_1=bh_2$ for some $h_1,h_2 \in H$. Thus $a=bh_2h_1^{-1}$.
Let $x\in Ha^{-1}$, then $x=ha^{-1}$ for some $h \in H$. From above we get,
\begin{align*}
x & = ha^{-1}\\
& = h(bh_2h_1^{-1})^{-1}\\
& = hh_1h_2^{-1}b^{-1}\\
& \in Hb^{-1}.
\end{align*}
This proves that $Ha^{-1} \subseteq Hb^{-1}$. Now you can do the reverse as well.
A: The line after "I am less sure of the reasoning" is correct.
If $b^{-1} a \in H$, then by the definition of a subgroup, $h(b^{-1} a) \in H$ for any $h \in H$. This implies $Hb^{-1} a \subseteq H$.
For the converse, if $h \in H$, then since $a^{-1} b \in H$ (which you've already shown), we have $h=(h a^{-1} b) b^{-1} a \in Hb^{-1} a$ so $H \subseteq Hb^{-1} a$.
A: You want to show that
$$
b^{-1}a \in H \implies Hb^{-1}=Ha^{-1}
$$
Now $b^{-1}a=h\in H$ implies $b^{-1}=ha^{-1}$, and hence $Hb^{-1}=Hha^{-1}=Ha^{-1}$.
