TITLE: Clarification on proof: Order of left cosets equal
QUESTION [9 upvotes]: There is a lemma that says that all left cosets $aH$ of a subgroup $H$ of a group $G$ have the same order.
The proof given is as follows...
The multiplication by $a \in G$ defines the map $H \rightarrow aH$ that sends $h\mapsto ah$. This map is bijective because its inverse is multiplication by $a^{-1}$.
I don't quite understand the proof. Why does having a bijective map mean that all sets of left cosets have the same order?
Thank you

REPLY [9 votes]: First, you should know that two sets $A,B$ have the same size (by
definition) if there is a bijection $f:A\to B$ (in this case there
is also a bijection $g:B\to A$).
You can understand why this is the definition in the case that $A,B$
are finite by using a drawing.
Second, $|aH|=|H|=|bH|$ (since the size of every coset is equal to
the size of $H$) and thus all cosets have the same cardinality.