A problem from Artin's algebra textbook

In summary: H∩K was of finite index in H, we get:[G:H][H:H∩K] cosets of H∩K in G in all, which would be finite.
  • #1
AbelAkil
9
0

Homework Statement


(a)Let H and K be subgroups of a group G. Prove that the intersection of xH and yK which are cosets of H and K is either empty or else is a coset of the subgroup H intersect K

(b) Prove that if H and K have finite index in G then the intersection of H and K also has finite index.

Homework Equations


The Attempt at a Solution


The intersection of xH and yK is a subgroup of both H and K, then how to continue?
 
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  • #2
AbelAkil said:

The Attempt at a Solution


The intersection of xH and yK is a subgroup of both H and K, then how to continue?
This is not true in general. If xH and yK are not subgroups, then neither contains the identity, so their intersection also doesn't contain the identiy. So it can't be a subgroup.

Moreover, in general [itex]xH \cap yK[/itex] isn't even a subSET of H or K. xH and H are disjoint unless [itex]x \in H[/itex]. Similarly for yK and K.
 
  • #3
jbunniii said:
This is not true in general. If xH and yK are not subgroups, then neither contains the identity, so their intersection also doesn't contain the identiy. So it can't be a subgroup.

Moreover, in general [itex]xH \cap yK[/itex] isn't even a subSET of H or K. xH and H are disjoint unless [itex]x \in H[/itex]. Similarly for yK and K.
Sorry, I made some mistakes when I wrote the post. In fact, I mean the intersection of H and K is a subgroup of both H and K...Could U give me some tips to prove it?
 
  • #4
If xH and yK have nonempty intersection, then there is an element g contained in both: [itex]g \in xH[/itex] and [itex]g \in yK[/itex].

The cosets of [itex]H \cap K[/itex] form a partition of G, so g is contained in exactly one such coset, call it [itex]a(H \cap K)[/itex].

If you can show that [itex]a(H \cap K)[/itex] is contained in both [itex]xH[/itex] and [itex]yK[/itex] then you're done.

Hint: both [itex]xH[/itex] and [itex]yK[/itex] are partitioned by cosets of [itex]H \cap K[/itex].
 
  • #5
jbunniii said:
If xH and yK have nonempty intersection, then there is an element g contained in both: [itex]g \in xH[/itex] and [itex]g \in yK[/itex].

The cosets of [itex]H \cap K[/itex] form a partition of G, so g is contained in exactly one such coset, call it [itex]a(H \cap K)[/itex].

If you can show that [itex]a(H \cap K)[/itex] is contained in both [itex]xH[/itex] and [itex]yK[/itex] then you're done.

Hint: both [itex]xH[/itex] and [itex]yK[/itex] are partitioned by cosets of [itex]H \cap K[/itex].
Yeah...I get it. Thanks very much. In addition, how to prove part (b), that is how can I show that both [itex]H[/itex] and [itex]K[/itex] are partitioned by finite cosets of [itex]H \cap K[/itex]... I appreciate your insightful answer!
 
  • #6
the index of H in G is the number of cosets of H.

if this number is finite, then if it just so happened that H∩K was of finite index in H, we get:

[G:H][H:H∩K] cosets of H∩K in G in all, which would be finite.

can you think of a way to show that [H:H∩K] ≤ [G:K]? perhaps you can think of an injection from left cosets of H∩K in H to left cosets of K in G?
 

1. What is "A problem from Artin's algebra textbook"?

"A problem from Artin's algebra textbook" refers to a specific problem or exercise found in the textbook "Algebra" by Michael Artin. This textbook is commonly used in undergraduate courses in mathematics, particularly in algebra and abstract algebra.

2. Is "A problem from Artin's algebra textbook" difficult?

That depends on your level of understanding and familiarity with the concepts covered in the textbook. Some problems may be more challenging than others, but with practice and a solid understanding of the material, they can be solved successfully.

3. How can I approach "A problem from Artin's algebra textbook"?

There is no one-size-fits-all approach to solving problems from Artin's textbook. However, some helpful tips include breaking the problem down into smaller parts, using examples to better understand the concepts, and seeking help from a professor or tutor if needed.

4. Can "A problem from Artin's algebra textbook" be solved using a calculator?

While some problems in Artin's textbook may involve numerical computations, the focus is on understanding and applying mathematical concepts rather than simply obtaining a numerical answer. Therefore, a calculator may not be necessary for most problems in the textbook.

5. Are there solutions available for "A problem from Artin's algebra textbook"?

Some solutions to problems in Artin's textbook may be available online or from other sources. However, it is important to try solving the problem on your own first, as this will help strengthen your understanding of the material. Additionally, consulting a solution should only be used as a last resort and not as a substitute for actively engaging with the material.

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