$|S_X|=|S_Y| \Leftrightarrow |X|=|Y|$

Question

Reading this problem I remembered trying to solve the following problem. For a set $A$, denote by $S_A=\{ f : A \to A | f \text{ is bijective }\}$. Denote by $|X|$ the cardinal number of $|X|$.

Prove that for two sets $X,Y$ we have $|X|=|Y| \Leftrightarrow |S_X|=|S_Y|$.

I didn't manage to solve the case where $X,Y$ are infinite, and don't even know how to start. I tried to represent $X$ as a subset of $Y$( if $|X|<|Y|$) and maybe find a permutation in $S_Y \setminus S_X$.

One cannot expect to be able to prove that if $|S_X|=|S_Y|$ then $|X|=|Y|$. — André Nicolas, Oct 18 '11 at 06:47
For the forward direction: if $|X|=|Y|$ then let $h\colon X\to Y$ be a bijection. Then I'm sure one can prove that $S_X = h^{-1}S_Yh = { h^{-1}\circ f\circ h\colon f\in S_Y }$, and hence $|S_X| = |S_Y|$. — Greg Martin, Oct 18 '11 at 06:52

score 3 · Accepted Answer · answered Oct 18 '11 at 06:55

3

First, note that your claim is false: if $|X|=0$, $|Y|=1$, then $|S_X|=|S_Y|=1$. If we exclude this case, then the claim is true for finite sets, since then we have $|S_X|=|X|!$.

If $X$ is infinite, then (assuming choice), $|S_X|=2^{|X|}$. So your claim becomes a claim about cardinal exponentiation. The generalized continuum hypothesis implies your claim, while the statement "$2^{\aleph_0}=2^{\aleph_1}$" refutes it. Both these statements are known to be independent of ZFC (assuming ZFC is consistent), so your claim is also independent of ZFC.

answered Oct 18 '11 at 06:55

Chris Eagle

33,306

2

On the other hand, the following is known: Every group isomorphism $S_X \to S_Y$ implies constructively that there is a bijection $X \to Y$. – Martin Brandenburg Oct 18 '11 at 07:25
@Martin Brandenburg: This is nice! Where can I find this? – Dec 27 '13 at 08:08
I don't know. For a proof, think about 2-cycles and 3-cycles. – Martin Brandenburg Dec 27 '13 at 10:18

$|S_X|=|S_Y| \Leftrightarrow |X|=|Y|$

1 Answers1