To prove this, we will show that (1) the collection of left cosets
covers G, and (2) left cosets are
disjoint.
1. For each
g∈G, we have that
g=g1H∈gH.
2. assume
g1=g2. Then for all
h∈H we have
g1h=g2h, due to the presence of inverses. Further,
g1H=g2H.