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

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Theorem[edit]

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[edit]

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


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

f is surjective[edit]

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


f is injective[edit]

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)[edit]

As f is surjective and injective,

6. f is a bijection from H to gH
7. Such bijection shows o(H) = o(gH)