Proving Group Equality for Normal Subgroups

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K is normal, which tells you that hkh-1is in K.In summary, the conversation discusses a proposed starting point for showing that if H and K are normal subgroups of a group G such that H∩K = {e}, then hk = kh for all h ∈ H and for all k ∈ K. It is suggested to show that hkh-1k-1 = e by proving that hkh-1k-1 ∈ H and hkh-1k-1 ∈ K. The process for showing both is similar, using the fact that H and K are normal subgroups.
  • #1
DanielThrice
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No. This isn't homework. And I think I am right there with this one.

I'm interested in the intersection of groups and what they equal, my professor proposed starting with something like this:

Show that if H and K are normal subgroups of a group G such that H∩K = {e}, then hk = kh for all h ∈ H and for all k ∈ K.

I've gotten this far:

kh-1k-1 = h(kh-1k-1) = (hkh-1)k-1 ∈ H∩K

Am I allowed to assume what I just did?
 
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  • #2
The first equality is wrong.
Essentially what you want to show is that hkh-1k-1=e.
But H∩K={e}, so just show that hkh-1k-1∈ H and hkh-1k-1∈ K.
 
  • #3
hkh-1k-1 = h(kh-1k-1; and H is normal, which tells you that kh-1k-1is in H. So hkh-1k-1 is the product of two elements of H and is therefore in H.
Is this correct? What about for K?
 
  • #4
Similair idea:

hkh-1k-1=(hkh-1)k-1
 
  • #5


I would say that you have made a valid assumption and have taken a logical approach to the problem. However, in order to fully prove the statement, you would need to provide a more formal and rigorous proof. This could involve using properties of normal subgroups, the definition of group equality, and other relevant theorems and definitions. It may also be helpful to provide a clear explanation of your thought process and reasoning behind each step. Overall, your approach seems to be on the right track and with some more work, you could potentially provide a solid proof for the given statement.
 

FAQ: Proving Group Equality for Normal Subgroups

Question 1: What is the concept of "Proving Group Equality for Normal Subgroups"?

"Proving Group Equality for Normal Subgroups" is a mathematical concept that involves showing that two subgroups of a larger group are equal or equivalent in terms of their elements and operations.

Question 2: Why is it important to prove group equality for normal subgroups?

Proving group equality for normal subgroups is important because it allows us to understand the structure and properties of a larger group by breaking it down into smaller, more manageable subgroups. It also helps us to identify any relationships or similarities between different subgroups.

Question 3: What are the key steps involved in proving group equality for normal subgroups?

The key steps involved in proving group equality for normal subgroups include showing that the subgroups have the same elements, that they have the same identity element, that they are closed under the group operation, and that they are inverses of each other.

Question 4: How do you prove normality of a subgroup in a larger group?

To prove normality of a subgroup in a larger group, we need to show that the subgroup is invariant under conjugation by any element in the larger group. This means that if we conjugate any element in the subgroup by an element in the larger group, the result will still be in the subgroup.

Question 5: Can you provide an example of "Proving Group Equality for Normal Subgroups"?

One example of proving group equality for normal subgroups is showing that the cyclic subgroups of a dihedral group are normal. This involves demonstrating that the cyclic subgroups are invariant under conjugation by any element in the dihedral group.

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