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There was a problem of finding out the number of permutations of order 2 in S4.

There are two cases.

case-1

permutation of single cycle of length 2.

case-2

permutations of two disjoint cycles each of length 2.

For case-1 total number of permutations will be $\frac{4P2}{2}=6$

and these permutations are $(1 2), (1 3),(1 4),(2 3),(2 4),(3 4)$

For case-2 total number of permutations will be $\frac{4P2}{2}\times\frac{2P2}{2}=6\times1=6$

But the permutations of two disjoint cycles each of length 2 are $(12)(34),(13)(24),(14)(23)$

this is clearly 3 and not 6.

I'm sure I am making some sort of mistake in finding the number of permutations in case-2.

Please help me out.

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  • $\begingroup$ I'm not sure I understand what makes you think that there are $6$ permutations which are products of two disjoint cycles. There are only $3$ of them. $\endgroup$ – Mark Apr 22 at 17:10
  • $\begingroup$ How did you come up with the formula for case-2? Can you extend on that? The correct answer is 3 $\endgroup$ – Ariel Bereslavsky Apr 22 at 17:12
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If I may repeat your argument back to you...

For case 2 (the case where the cycle structure is $(..)(..)$), you say that the total number of permutations is $$ \frac{4!}{2!2!} \times \frac{2!}{2!2!} = 6.$$

This is made up of:

  • $\frac{4!}{2!2!} = 6$, the number of ways of choosing the first $2$-cycle.
  • $\frac{2!}{2!2!} = 1$, the number of ways of choosing the second $2$-cycle, once the first $2$-cycle has already been chosen.

These $6$ permutations are: $$ (12)(34), \ (13)(24), \ (14)(23), \ (23)(14), \ (24)(13), \ (34)(12).$$

However, there is a subtlety!

  • $(12)(34)$ is really the same thing as $(34)(12)$.
  • $(13)(24)$ is really the same thing as $(24)(13)$.
  • $(14)(23)$ is really the same thing as $(23)(14)$.

In other words, we have double-counted.

To make up for the double-counting, we divide the whole answer by $2$, giving

$$ \frac{4!}{2!2!} \times \frac{2!}{2!2!} \times \frac 1 2 = 3,$$

which is now the correct answer.

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