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

Theorem

Let H₁, H₂, ... H_n 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

$(H_1\cap H_2)\subseteq G$

1. $H_1\subseteq G$	H1 is [[../Definition of a Subgroup|subgroup]] of G
2. $H_2\subseteq G$	H2 is [[../Definition of a Subgroup|subgroup]] of G
3. $(H_1\cap H_2)\subseteq G$	1. and 2.

$H_1\cap H_2$ with $\ast$ is a Group

Closure

4. Choose $x,y\in (H_1\cap H_2)$
5. $x\ast y\in H_1$	[[../../Group/Definition of a Group/Definition of Closure|closure]] of H1
6. $x\ast y\in H_2$	[[../../Group/Definition of a Group/Definition of Closure|closure]] of H2
7. $x\ast y\in (H_1\cap H_2)$	5. and 6.

Associativity

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

11. $e_G\in H_1$ and $e_G\in H_2$	Subgroup H1 and H2 [[../Subgroup Inherits Identity|inherit identity]] from G
12. $\mathrm{\forall }g\in G:e_G\ast g=g\ast e_G=g$	eG is [[../../Group/Definition of a Group/Definition of Identity|identity]] of G,
13. $\mathrm{\forall }g\in (H_1\cap H_2):e_G\ast g=g\ast e_G=g$	$(H_1\cap H_2)\subseteq G$ and 9.
14. $(H_1\cap H_2)$ has identity eG	[[../../Group/Definition of a Group/Definition of Closure|definition of identity]]

Inverse

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. $g_{H1}^{-1}\in G$	$H_1\subseteq G$
19. $g_{H1}^{-1}\ast g=g\ast g_{H1}^{-1}=e_G$	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. $g^{-1}=g_{H1}^{-1}=g_{H2}^{-1}\in (H_1\cap H_2)$
24. g has inverse g−1 in $(H_1\cap H_2)$