# number of permutation of S4 as product of two disjoint cycles each of length 2

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.

• 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. – Mark Apr 22 at 17:10
• How did you come up with the formula for case-2? Can you extend on that? The correct answer is 3 – Ariel Bereslavsky Apr 22 at 17:12

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 $$\frac{4!}{2!2!} \times \frac{2!}{2!2!} = 6.$$

• $$\frac{4!}{2!2!} = 6$$, 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.