Abstract Algebra/Group Theory/Subgroup/Intersection of Subgroups is a Subgroup

= Theorem = Let H1, H2, ... Hn be subgroups of Group G with operation $$\ast$$


 * $$ H_1 \cap H_2 \cap \cdots \cap H_n $$ with $$\ast$$ is a subgroup of Group G

= Proof =

= $$\color{RawSienna}(H_1 \cap H_2) \subseteq G$$ =


 * {| Style = "width:60%"

|-    |-     |-
 * 1. $$ H_1 \subseteq G$$
 * H1 is subgroup of G
 * 2. $$ H_2 \subseteq G$$
 * H2 is subgroup of G
 * 3. $$(H_1 \cap H_2) \subseteq G$$
 * 1. and 2.
 * }

= $$\color{RawSienna}H_1 \cap H_2$$ with $$\color{RawSienna}\ast$$ is a Group =

Closure

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 * 4. Choose $$x,y \in (H_1 \cap H_2)$$
 * 5. $$ x \ast y \in H_1 $$
 * closure of H1
 * 6. $$ x \ast y \in H_2 $$
 * closure of H2
 * 7. $$ x \ast y \in (H_1 \cap H_2)$$
 * 5. and 6.
 * }

Associativity

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 * 8. $$ \ast $$ is associative on G.
 * Group G's operation is $$ \ast $$
 * 9. $$(H_1 \cap H_2) \subseteq G$$
 * 3.
 * 10. $$\ast$$ is associative on $$(H_1 \cap H_2)$$
 * 8. and 9.
 * }

Identity

 * {| Style = "width:75%"

|-
 * 11. $$ e_{G} \in H_1$$ and $$ e_{G} \in H_2$$

|-    |-     |-
 * Subgroup H1 and H2 inherit identity from G
 * 12. $$ \forall g \in G: e_{G} \ast g = g \ast e_{G} = g $$
 * eG is identity of G,
 * 13. $$ \forall \; g \in (H_1 \cap H_2): e_{G} \ast g = g \ast e_{G} = g $$
 * $$(H_1 \cap H_2) \subseteq G$$ and 9.
 * 14. $$(H_1 \cap H_2)$$ has identity eG
 * definition of identity
 * }

Inverse

 * {| Style = "width: 75%"

|-    |-     |-     |-     |-     |-     |-     |-     |-     |-     |-
 * 15. Choose $$ g \in (H_1 \cap H_2) \subseteq G $$
 * 16. $$ g \in H_1$$, $$ g \in H_2$$, and $$ g \in G$$
 * 17. gH1−1 in H1, and gH2−1 in H2.
 * G, H1, and H2 are groups
 * 18. $$ g^{-1}_{H1} \in G$$
 * $$ H_{1} \subseteq G $$
 * 19. $$ g^{-1}_{H1} \ast g = g \ast g^{-1}_{H1} = e_{G} $$
 * G and H1 shares identity e
 * 20. gH1−1 is inverse of g in G
 * 19. and definition of inverse
 * 21. Let gG−1 be inverse of g has  in G
 * 22. gG−1 = gH1−1
 * inverse is unique
 * 22. gG−1 = gH2−1
 * similar to 21.
 * 23. $$ g^{-1} = g^{-1}_{H1} = g^{-1}_{H2} \in (H_1 \cap H_2) $$
 * 24. g has inverse g−1 in $$(H_1 \cap H_2)$$
 * }