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

Theorem

Let g be any element of group G.

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

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

o(H) = o(gH)

Proof

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

0. Define ${\begin{aligned}f\colon H&\to gH\\h&\mapsto gh\end{aligned}}$

f is surjective

1. f is surjective by definition of gH and f.

f is injective

2. Choose ${\color {Blue}h_{1}},{\color {OliveGreen}h_{2}}\in H$ such that $f({\color {Blue}h_{1}})=f({\color {OliveGreen}h_{2}})$
3. $g{\color {Blue}h_{1}}=g{\color {OliveGreen}h_{2}}$	0.
4. $g,{\color {Blue}h_{1}},{\color {OliveGreen}h_{2}}\in G$	${\color {Blue}h_{1}},{\color {OliveGreen}h_{2}}\in H$, and subgroup $H\subseteq G$
5. ${\color {Blue}h_{1}}={\color {OliveGreen}h_{2}}$	3. and cancelation justified by 4 on G

o(H) = o(gH)

As f is surjective and injective,

6. f is a bijection from H to gH

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