TITLE: Equivalence Relation in a Subgroup
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: $a \sim b$ 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 $|H|$.