# Abstract Algebra/Group Theory/Subgroup/Coset/a Subgroup and its Cosets have Equal Orders

# Theorem

editLet *g* be any element of group G.

Let H be a subgroup of G. Let o(H) be order of group H.

Let *g*H be coset of H by *g*. Let o(gH) be order of *g*H

- o(H) = o(gH)

# Proof

editOverview: A bijection between H and gH would show their orders are equal.

- 0. Define

*f* is surjective

edit
- 1.
*f*is surjective by definition of*g*H and*f*.

*f* is injective

edit
2. Choose such that - 3.

0. - 4.

, and subgroup - 5.

3. and cancelation justified by 4 on G

## o(H) = o(*g*H)

edit
As *f* is surjective and injective,

- 6.
*f*is a bijection from H to*g*H

- 6.

- 7. Such bijection shows o(H) = o(
*g*H)

- 7. Such bijection shows o(H) = o(