Equivalent Order of Cosets

Theorem

Given a group G with subgroup H, the order of all cosets of H in G (both left and right) are equal, which is in turn equal to the order of H itself. That is, for all g∈G

|gH|=|H|=|Hg|

and hence

|gH|=|g′H|.
Proof

This can be proven by showing that the map f:H→gH given by

h↦gh

is bijective.

Here we prove this from scratch, although this fact should be obvious given that multiplication by g on the left is invertible through multiplication by g−1 on the left.

For injectivity we have

f(h)=f(h′)⟺gh=gh′⟺g−1gh=g−1gh′⟺h=h′

and for surjectivity we have for any gh∈gH we have

f(h)=gh.