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Let $A = \{1,2,3,4\}$. Number of functions $f :A \to A$, such that $f(f(x))=x, \forall x \in A$ , is.

For this , what I thought was 2 cases are possible. Either an element maps to itself or 2 elements map to each other.

Eg either $1 \to 1$ or $1 \to 2$ and $2 \to 1.$

So for this I can group the elements in groups of 2 in $^4C_2 $ways. Then say I group $(a,b)$ and$ (c,d)$ . Then $ a$ can either map to $a$ or $b$. The other element from that group will automatically be defined. Similarly for $(c,d)$

So the answer should be $^4C_2× 2×2 = 24$ . But the answer is

$13$

Please help.

Mr.HiggsBoson
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  • You're looking for elements $\sigma\in S_4$ such that $\sigma^2=id$. These are precisely product of disjoint 2-cycles. – Rodrigo Dias Jul 30 '19 at 20:02
  • An alternative interpretation yielding the same answer would be to consider the number of fixed points of $f$ and realizing it is necessarily even. – Thorgott Jul 30 '19 at 20:04
  • Can you please tell , where did I go wrong , where have I counted the the same multiple times ?? . How can I make sure to not make such errorsi n the future ?? – Mr.HiggsBoson Jul 30 '19 at 20:11
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    For grouping them into pairs, choosing (1,3) and choosing (2,4) give you the same choice, so there are only three ways of doing this grouping, not six ways. (Choose what to pair with 1, and that determines it.) Then for each of the three ways of dividing into pairs, if you decide to send each element to itself, you end up with the same function, so you are counting that function three times. – John Palmieri Jul 30 '19 at 20:29
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    The answer in the text is wrong. There are ten involutions of a set with four elements. – N. F. Taussig Jul 30 '19 at 20:43

2 Answers2

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Case 1: There are 4 fixed points. The only such involution is the identity.

Case 2: There are 2 fixed points. In this case, once 2 elements are fixed, the remaining 2 elements must map to each other. Hence, the number of involutions with 2 fixed points is $\binom{4}{2} = 6$.

Case 3: The involution is a derangement (fixed-point-free). In this case, the number 1 must map to 2, 3, or 4, and whatever 1 maps to must be mapped to 1. Then, the remaining 2 elements must map to each other. Hence, the number of involutions without any fixed points is 3.

In total, there are 10 involutions on a 4-element set like $A = \{1, 2, 3, 4\}$.

Geoffrey Trang
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  • Can you please tell , where did I go wrong , where have I counted the the same multiple times ?? . How can I make sure to not make such errorsi n the future ?? – Mr.HiggsBoson Jul 30 '19 at 20:15
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Jack Schmidt's answer here can help you figure out the elements of $S_4$.

https://math.stackexchange.com/a/379912/399263

Elements of order $2$ are:

  • the $6$ transpositions $(1,2)\quad (1,3)\quad (1,4)\quad (2,3)\quad (2,4)\quad (3,4)$
  • the $3$ products of disjoint transpositions $(1,2)(3,4)\quad (1,3)(2,4)\quad (1,4)(2,3)$

Note that since the transpositions are disjoint, $(i,j)(k,l)$ is the same as $(k,l)(i,j)$, do not count double transpositions twice.

And you have to add the identity to these, giving a total of $6+3+1=10$ involutions.

zwim
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