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