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A question came up into mind.

Prove that for every nonempty set $X$ an operation can be suggested such that $X$ would be a group with that operation. For example, it is obvious for finite and countable sets: $(\mathbb{Z}_n,+), (\mathbb{Q}, +)$. Also, it can be done for all sets of form $X=2^L$, as $(X, \Delta)$, which is symmetric difference on subsets of $L$.

So is seems that the question is reduced to (1) sets which are high in hierarchy of cardinals (not of form $2^L$) and (2) sets which do not exist assuming continuum hypothesis (between $\mathbb{R}$ and $2^\mathbb{R}$ for instance).
Axiom of choice is given.

Or maybe intuition is wrong and for some sets it cannot be done, then a proof of existance of such or a single example would be nice. Thank you in advance.

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    $\begingroup$ It is a fact that "putting a group structure can be put on any set" is equivalent to the axiom of choice. This MO post gives a discussion of that. $\endgroup$
    – user29123
    Dec 19, 2015 at 0:02
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    $\begingroup$ @PaulPlummer Someone should write a book on statements that are not equivalent to AC. If they can think of any, I mean... $\endgroup$ Dec 19, 2015 at 0:12
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    $\begingroup$ This is a statement that is not equivalent to the axiom of choice. $\endgroup$ Dec 19, 2015 at 0:22
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    $\begingroup$ @JoshuaMeyers Thanks... $\endgroup$ Dec 19, 2015 at 0:24

3 Answers 3

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If it can be done for one set it can be done for every other set of the same cardinality (just use a bijection to copy the structure).

So just take a set $X$. Take the set $\Omega$ of all finite subsets of $X$, it is of the same cardinality as $X$ (we need axiom of choice for this).

Now check $\Delta$ ( symmetric difference) makes $\Omega$ into a group, we are done.

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  • $\begingroup$ To be pedantic, this only works as stated for infinite sets: for finite $X$, $\Omega$ has strictly higher cardinality than $X$. Also, you might want to explicitly note that $\Delta$ denotes (as I assume) the symmetric set difference operator. $\endgroup$ Dec 19, 2015 at 2:56
  • $\begingroup$ Yeah, although the OP specifically uses it in his post. $\endgroup$
    – Asinomás
    Dec 19, 2015 at 3:00
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    $\begingroup$ Note that the place where choice creeps in is the statement "the set of all finite subsets is of the same cardinality" - this need not be true in general without choice. $\endgroup$ Dec 19, 2015 at 3:04
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The free group (or the free Abelian group, etc.) generated by an infinite set $X$ will have the same cardinality as $X.$ Therefore, every nonempty set is the underlying set of a group.

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If $\kappa$ is an infinite cardinal then the cardinality of the collection of all finite subsets of $\kappa$ is again $\kappa$. So the direct sum of $\kappa$ copies of $\Bbb Z_2$ is a group of cardinality $\kappa$.

(Direct sum, not product: The space of all functions $f:\kappa\to\Bbb Z_2$ such that $f(\alpha)=0$ for all but finitely many $\alpha$.)

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