Lemma to prove Lagrange's Theorem In the second part of a lemma used to prove Lagrange's Theorem (subgroups of a finite group g), it is stated that 
$ Ha \mapsto a^{-1}H $ is a bijection $Ha \mapsto bH $ where $a,b \in G $
In the proof it says that $ Ha = Hb$ iff $ab^{-1} \in H$ iff $a^{-1}H = b^{-1}H $. Up until this point I understand. Next it states that this shows that the map is well defined, one-to-one and onto. Why is this true? Thanks 
 A: Let $\Gamma$ be your map.


*

*Well defined: You must show that $Ha_1=Ha_2\implies a_1^{-1}H=a_2^{-1}H$. In fact, the first equality is equivalent to $a_1a^{-1}_2\in H$, which implies the second by the previous observation.

*Surjective: $bH=\Gamma(Hb^{-1})$.

*Injective: $\Gamma(Ha_1)=\Gamma(Ha_2)\iff a_1^{-1}H=a_2^{-1}H\iff a_1a_2^{-1}\in H\iff Ha_1=Ha_2$
A: If you understand the equalities and the iff's, then the one-to-one, onto, and well-defined all come from the iff's.
Here's how to check injectivity:  Suppose that $Ha\mapsto cH$ and $Hb\mapsto cH$.  Then $a^{-1}H=cH=b^{-1}H$ (by the definition of the map).  By the iff's, this is true iff $ac\in H$ and $c^{-1}b^{-1}\in H$ (it might help to briefly write $c$ as $(c^{-1})^{-1}$ so that you can match up the forms).  By multiplying these, we have that $acc^{-1}b^{-1}=ab^{-1}\in H$ (since a product of two elements of $H$ is in $H$).  Which, by the iffs, implies that $aH=bH$.
A: It's well defined, because if you take two representatives $a$ and $b$ of the same class ($Ha=Hb$), their images are the same ($a^{-1}H=b^{-1}H$).
It is one-to-one because $a^{-1}H=b^{-1}H$ implies $Ha=Hb$.
It is onto because any left class $aH$ is the image of the right class $Ha^{-1}$, since $\bigl(a^{-1}\bigr)^{-1}=a$.
