Prove that the intersection of any two subgroups of a group is again a subgroup?

Answer 1

Been a while since I took abstract algebra so I may not remember all the axioms that define a group, but I'll make a start based on what I remember and you can figure the rest. Note -- this is NOT a formal proof, only a clue to get you headed in the right direction.

First, any subgroup is a group in its own right, having commutative and associative addition and multiplication, closure of addition and multiplication, and identity elements for addition and multiplication.

If you take two subgroups and intersect them, you pick up all that they have in common. This would pick up the identity elements for both addition and multiplication. So that's there.

Closure is the trickiest part. But it seems to me that if you intersect two subgroups, again picking up all that they have in common, you pick up the results of combining these elements via addition and multiplication. You'l have to work on this part some, so you can have closure.

Your operations would be inherited from the group that your subgroups are taken from, so that shouldn't be too much of a problem.

Good luck!

Answer 2

Mostly.....you just say "inherited from the group." The only really key notion is closure. If A and B are subgroups, and x and y are elements in A intersect B, then x and y are in A and since A is a subgroup, x*y is in A. Similarly, x,y in B means x*y in B, so x*y in A intersect B. So A inersect B is closed under the group law. Iguess you also ...yeah, the other post, go through the whole list...the only freebie is associativity, you do "prove" that but it comes from the group. But that is the idea. In each individual one so in the intersection.

Answer 3

A set H is a subgroup of a group G if a subset of G and is a group using the operation (*) defined on G. In other words, H is a subgroup of (G, *) if the restriction of * to H is a group operation on H.

So, two subsets in G called H1 and H2 are given to be subgroups. It follows directly that all elements of H1 and H2 are also elements of G, therefore the intersection of H1 and H2 must be in G, satisfying the first criterion of the intesection being a subset of G.

To satisfy the second criterion, it is necessary to prove that the restriction of * to H1 int. H2 is a group operation on H1 int. H2. Since all elements of G already satisfy that criterion, and since H1 and H2 are in G, then the intersection is necessarily in G. Thus the group operation must be valid for H1 int H2.

This seems trivial, but I think that it makes sense that it follows from the definitions. (I believe this is the same argument that a_math_guru is making.)

Answer 4

intersection of the subgroups will have elements in the main group.
so that will also be a sub group
example
let A={1,2,3,4,5,6,7,8,9,10}
B={1,3,4,5,6}
C={2,4,6,8}
B and Care subgroups of A
B int C is {4,6}
which is again a subgroup of A

Answer 5

go through the axioms of what it means to be a group, and check that they hold for this particular subset you are interested in.