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

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Theorem[edit]

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[edit]

\color{RawSienna}(H_1 \cap H_2) \subseteq G[edit]

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 [edit]

Closure[edit]

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[edit]

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[edit]

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[edit]

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)