Let 𝐺 be a group, 𝑔1,𝑔2∈𝐺, and 𝐻≤𝐺 be a subgroup.
#m/thm/group Then the left cosets𝑔1𝐻,𝑔2𝐻 are
identical iff 𝑔−11𝑔2∈𝐻
disjoint otherwise
Proof
Let 𝑔1,𝑔2∈𝐺, and 𝐻⊆𝐺 be a subgroup.
Due to basic Properties,
𝑔−11𝑔2∈𝐻⟺𝑔−11𝑔2𝐻=𝐻⟺𝑔1𝐻=𝑔2𝐻
Next assume there exist ℎ1,ℎ2∈𝐻 such that 𝑔1ℎ1=𝑔2ℎ2,
i.e. 𝑔1𝐻 and 𝑔2𝐻 have a common element.
Then 𝑔1=𝑔2ℎ2ℎ−11,
whence 𝑔1𝐻=𝑔2ℎ2ℎ−11𝐻
and since ℎ2ℎ−11∈𝐻,
it follows 𝑔1𝐻=𝑔2𝐻
and thus 𝑔−11𝑔2∈𝐻.
Hence is 𝑔−11𝑔2∉𝐻,
𝑔1𝐻 and 𝑔2𝐻 can share no common element.