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.]]>

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

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.

]]>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.]]>