Given an infinite set $A$  does the cardinality of $A$ equal to the cardinality of $A^2$?
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4Yes, assuming the axiom of choice. – Brian M. Scott Aug 09 '12 at 13:38

Do you mean *group*, which is a set *with* a binary operation; or just a *set*? – Asaf Karagila Aug 09 '12 at 14:17

2See [here](http://math.stackexchange.com/a/54904/). – Martin Sleziak Aug 09 '12 at 14:19

My bad  A is a set  not group. – poli Aug 09 '12 at 14:42

As a simple plausibility argument, consider the case where $A=\mathbb{R}$. Then clearly we can take the decimal expansions of two real numbers and combine them into one decimal expansion, like shuffling a pack of cards. It's also pretty easy to do an explicit construction for $\mathbb{Z}^S$, where $S$ is any set. Since we think of big sets as being built from powersets of smaller sets, this makes it pretty plausible that it holds in general. – Aug 09 '12 at 19:03
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Zermelo proved that every wellordered infinite set has this property, so if we assume the axiom of choice then the answer is yes. In fact the axiom of choice is equivalent to the assertion that for every infinite set $A$, $A$ and $A^2$ are equinumerous.
 The proof of Zermelo's theorem can be found here: About a paper of Zermelo
 The proof for the reverse implication can be found here: For every infinite $S$, $S=S\times S$ implies the Axiom of choice
However if the axiom of choice is negated there are sets whose cardinality is strictly less of that of their square. We even know how to construct such set:
Suppose that $X$ cannot be wellordered (such set exists, since we assume the axiom of choice is false, and therefore the wellordering principle is false). Let $\kappa$ be an ordinal such that there is no $f\colon\kappa\to X$ which is injective.
In this case $A=X\cup\kappa$ has the property that $A^2$ is strictly larger than $A$.
Asaf Karagila
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You proved that $S=S^2$ implies AC, but the person asking the question is interested in proving $S=S^2$, presumably in ZFC. – Aug 09 '12 at 18:48

@Ben: You're right, I was going to edit in the refernce to my answer from Martin's comment, but I forgot. I'll do that now. – Asaf Karagila Aug 09 '12 at 18:49

The story about why Comptes Rendu rejected Tarski's paper is hilarious. I find the theorem surprising in that it implies full AC rather than some restricted version of choice. But I suppose the sets that violate $A^2=A$ are not explicitly constructible...? – Aug 09 '12 at 19:26

You mean Lebesgue. That depends on what you mean constructible. I gave you a way to construct such set given a set which is not wellorderable. If the real numbers fails to be wellorderable then you can use them as the starting set. – Asaf Karagila Aug 09 '12 at 19:54