Equivalence Relation in a Subgroup Let $G$ be a finite group and $H$ a subgroup. Define a relation on $G$ by $$a\sim b\iff b^{-1}a \in H.$$
(0.) Show that this is an equivalence relation.
(i) Prove that for this relation $$[a] = \{ah: h\in H\}.$$
(ii) Prove that the cardinality $[a]$ equals the cardinality of $|H|$. 
(iii) Conclude that $|H|$ divides $|G|$.
Attempts/Ideas:
(0):  $a\sim a$ because $a^{-1}a\in H$, since $e$ is in $H$.
$b\sim a$ : Show $a^{-1}b = b^{-1}a$
$a\sim b$ and $b \sim c$ imply $a\sim c$ : Show $(c^{-1}b)(b^{-1}a)\in H$.
(i) I'm not sure where to start here, except maybe with an idea from (0.). 
(ii) How to show a 1-1 correspondence, if that should work.
(iv.) Use partitions of $G$ to say that $H$ divides $G$. 
Thx!
Answers and help would be appreciated!
 A: $a \sim b$ is an equivalence relation because:


*

*reflexivity: $a^{-1} a = 1 \in H$ because groups contain identity.

*symmetry: $b^{-1} a \in H$ implies $a^{-1} b = (b^{-1} a)^{-1} \in H$ because groups contain inverses.

*transitivity: $b^{-1} a, c^{-1} b \in H$ implies $c^{-1} a = (c^{-1} b) (b^{-1} a) \in H$ because groups are closed under multiplication.



Let us compute an equivalence class $$[a] = \{ b \in H | a \sim b \}$$ this is the same as $\{ b \in H | a^{-1} b \in H \}$ by symmetry and $a^{-1} b \in H$ is equivalent to $b \in a H$ therefore  $$[a] = a H$$ there are called cosets.

The cardinality of $[a]$ is equal to the cardinality of $H$ because the map $h \mapsto ah : H \to [a]$ is a bijection. The reason it is a bijection is because multiplication by a group element is invertible.

Hint For any $a,b \in G$ either $[a] = [b]$ or $[a] \cap [b] = \{\}$.
This lets you prove $|H|$ divides $|G|$ since you can partition $G$ into classes of size $|H|$.
A: (0.) Fine for reflexivity.  For symmetry, your idea is not going to work, because it's not always the case that $a^{-1}b = b^{-1}a$.  Luckily, you don't need to show that.  You only need $b^{-1}a$ to be in $H$ whenever $a^{-1}b$ is in $H$. 
For transitivity, your idea will work.  Do you see why?
(i)  What do you need to do in general to show that two sets are equal?
The other parts should probably be fine once you've got the first two sorted.
