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

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# 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)