Abelian Group/Subgroups of Power Algebra

  • Thread starter Thread starter aeronautical
  • Start date Start date
  • Tags Tags
    Algebra Power
Click For Summary
SUMMARY

The discussion focuses on the properties of the power algebra P(X) of a set X with three elements, specifically its structure as an Abelian group under the symmetric difference operator, denoted as delta. Participants emphasize that the power set contains eight elements, allowing for the identification of subgroups of orders 1, 2, 4, and 8. The conversation highlights the necessity of understanding group theory fundamentals, such as the identity element and Lagrange's theorem, to effectively determine the subgroups of P(X).

PREREQUISITES
  • Understanding of group theory concepts, including groups and subgroups
  • Familiarity with Lagrange's theorem
  • Knowledge of symmetric difference operation in set theory
  • Ability to construct and interpret Cayley tables
NEXT STEPS
  • Study the definition and properties of Abelian groups
  • Learn about the symmetric difference operation in detail
  • Explore Lagrange's theorem and its implications for subgroup orders
  • Practice constructing Cayley tables for small groups
USEFUL FOR

Students of abstract algebra, mathematicians interested in group theory, and anyone seeking to understand the structure of power algebras and their subgroups.

aeronautical
Messages
33
Reaction score
0

Homework Statement



Let X be a set with exact three elements. Then its power algebra P(x), is an Abelian group with symmetrical difference operator delta.
A subset H of P(x) is a subgroup if, for all A, B in H, A deltaB in H.
Find all possible subgroups of P(x).

The Attempt at a Solution



I have no clue how to begin solving this problem. But I know that the power set of any set X becomes an Abelian group if we use the symmetric difference as operation. How can I show this?
 
Physics news on Phys.org
The power set has only 8 elements. So you can just do it by exhaustion (and a little common sense: subgroups of order 2 are trivial to find, so that just leaves order 4, and there are only two groups of order 4 up to isomorphism).
 
matt grime said:
The power set has only 8 elements. So you can just do it by exhaustion (and a little common sense: subgroups of order 2 are trivial to find, so that just leaves order 4, and there are only two groups of order 4 up to isomorphism).

Can you please refer me to any books to read about this topic? I do not really understand what you mean. Can you please elaborate? Thanks...
 
Start with subgroups of order 2. What is a (sub)group of order 2? It is a set {e,g} with e the identity (which is what in this case?) and g satisfying what?

Just to check where the problem lies: do you understand why the only subgroups have orders 1,2,4, and 8?
 
Actually with this problem... I am not even sure where to start... I don't know what the subgroup of order 2 is. I have nothing specified apart from the problem statement, which I agree is rather confusing. I do not understand why the only subgroups have the orders 1, 2, 4 and 8?

I read that the identity of the subgroup is the identity of the group. But with this proof... I am totally lost of where I should start. Could you please guide me?
 
You should start with the basics of group theory: the definition of a group, subgroup, Lagrange's theorem, for example. If you don't know what a subgroup of order 2 is, then you'll never find one.

I don't think there is anything confusing about the question if you know what the definitions are.

First, identify the identity element in the group. Actually, first, what are the elements of the group? What is the Cayley table (or multiplication table) for the group? If a set with 3 elements is too big, try a set with 1 element, then 2 elements. Just take two elements in the group and compose them, see what happens.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
574
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K