Abstract Algebra/Group Theory/Subgroup/Subgroup Inherits Identity

= Theorem = Let H be subgroup of Group G. Let $$ \ast $$ be the binary operation of both H and G


 * H and G shares identity

= Proof =
 * {| Style="width:60%"

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 * 0. Let eH, eG be identities of H and G respectively.
 * 1. $$ {\color{OliveGreen}e_{H}} \ast {\color{OliveGreen}e_{H}} = {\color{OliveGreen}e_{H}} $$
 * 1. $$ {\color{OliveGreen}e_{H}} \ast {\color{OliveGreen}e_{H}} = {\color{OliveGreen}e_{H}} $$

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 * eH is identity of H (usage 1, 3)
 * 2. $$ {\color{OliveGreen}e_{H}} \in H $$
 * 2. $$ {\color{OliveGreen}e_{H}} \in H $$

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 * eH is identity of H (usage 1)
 * 3. $$ H \subseteq G $$
 * 3. $$ H \subseteq G $$

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 * H is subgroup of G
 * 4. $$ {\color{OliveGreen}e_{H}} \in G $$
 * 4. $$ {\color{OliveGreen}e_{H}} \in G $$

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 * 2. and 3.
 * 5. $$ {\color{OliveGreen}e_{H}} \ast {\color{Blue}e_{G}} = {\color{OliveGreen}e_{H}} $$
 * 5. $$ {\color{OliveGreen}e_{H}} \ast {\color{Blue}e_{G}} = {\color{OliveGreen}e_{H}} $$

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 * 4. and eG is identity of G (usage 3)
 * 6. $$ {\color{OliveGreen}e_{H}} \ast {\color{Blue}e_{G}} = {\color{OliveGreen}e_{H}} \ast {\color{OliveGreen}e_{H}} $$
 * 6. $$ {\color{OliveGreen}e_{H}} \ast {\color{Blue}e_{G}} = {\color{OliveGreen}e_{H}} \ast {\color{OliveGreen}e_{H}} $$

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 * 1. and 5.
 * 7. $$ {\color{Blue}e_{G}} = {\color{OliveGreen}e_{H}} $$
 * 7. $$ {\color{Blue}e_{G}} = {\color{OliveGreen}e_{H}} $$


 * cancellation on group G
 * }

= Usages =
 * 1)  If H is subgroup of group G, identity of G is identity of H.
 * 2)  If H is subgroup of group G, identity of G is in H.