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{align}

f\colon H &\to gH \\ h &\mapsto gh \end{align}$$

f is surjective

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

f is injective

 * {| Style = "width:70%"

|-    |-
 * 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. $$ g{\color{Blue}h_1} = g{\color{OliveGreen}h_2} $$
 * 3. $$ g{\color{Blue}h_1} = g{\color{OliveGreen}h_2} $$

|-
 * 0.
 * 4. $$g, {\color{Blue}h_1}, {\color{OliveGreen}h_2} \in G$$
 * 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. $$ {\color{Blue}h_1} = {\color{OliveGreen}h_2}$$
 * 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)