Abelian Group/Subgroups of Power Algebra

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Homework Help Overview

The problem involves the power algebra of a set with three elements, specifically examining the structure of its power set as an Abelian group under the symmetric difference operation. The task is to identify all possible subgroups of this group.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the feasibility of finding subgroups by exhaustion due to the limited number of elements in the power set. Questions arise regarding the nature of subgroups, particularly those of order 2, and the identity element within the group.

Discussion Status

The discussion is ongoing, with participants exploring foundational concepts of group theory and subgroup identification. Some guidance has been offered regarding the importance of understanding group definitions and the identity element, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants express confusion regarding the problem statement and the specific properties of subgroups, particularly the orders of potential subgroups. There is a noted lack of clarity on foundational group theory concepts among some participants.

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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?
 
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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.
 

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