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**Seetha Rama Raju Sanapala****Member**- Registered: 2016-10-23
- Posts: 4

I want to know if we can have a group of at least 2 elements with sets as elements and some set operation (union, intersection or some other) as the binary operation.

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**Lehona****Member**- Registered: 2016-10-22
- Posts: 16

I could only think of the complement, but that doesn't fulfill the associativity requirement.

You could argue that in modular arithmetic, every element (e.g. of (Z_7, +)) is actually a set, but the operation wouldn't be one of the usual set operations.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,869

hi Seetha Rama Raju Sanapala

Welcome to the forum.

I am replacing an earlier post with this version. My apologies to anyone who spent time on my previous version. Hopefully, I've got it right now.

Note: both intersection and union are associative.

I'll consider intersection first.

If intersection is the binary operation for a group, there must be an identity set; let's call it I.

Also we want another member; let's call it A.

For a group we require inverses; let's call the inverse of A, B. So

This means that I is contained in A {and also in B).

But I is the identity so

So A is contained in I.

So A is contained in I and I is contained in A. This means that A = I.

So every element of the group is I. Conclusion: no such groups with 2 or more members.

The case for union follows similar lines.

So A is contained in I

But I is an identity so

So I is contained in A. etc etc.

Bob

ps. Sorry but your full name is too long to fit the blue space above.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Seetha Rama Raju Sanapala****Member**- Registered: 2016-10-23
- Posts: 4

Dear Bob Bundy,

Thanks a lot for the clear proof of impossibility of union and intersection as the operation of a group. But there can be other set operations like difference, symmetric difference, complement of sets etc - any thing based on the elements of the sets . Can we show that they are all ruled out?

Thanks,

Seetha Rama Raju Sanapala

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**Lehona****Member**- Registered: 2016-10-22
- Posts: 16

Symmetric Difference should be doable, right? A o B = B o A, ∅ is the neutral element and A^(-1) = A.

Closure should be achievable, too, right? Let's say A o B = C, then A o C = B -> Whatever you do, you still land in the group. It's late at night here, so please someone correct me if I'm wrong, but it sounds plausbile in my head.

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