# Permutation groups

1. Nov 14, 2004

### astronut24

what is the number of elements of order 5 in the permutaion group S7??
so what we're concerned with here is, after decompositon into disjoint cycles the l.c.m of the lengths must be 5. since 5 is a prime, the only possible way we could get 5 as l.c.m would be to fix ANY 2 elements amongst the 7 to themselves....so we end up getting 2 cycles of length 1 each. the remaining five elements can be arranged in 4! ways...
so, the answer is 7C2 * 4! = 21*24 = 504.
but unfortunately, this answer is WRONG!!
CAN ANYBODY TELL ME WHY?! PLEEEEASE HELP!

2. Nov 14, 2004

### Galileo

Since 5 is prime, you MUST have cycle of length 5.
Then there could be a 2-cycle in addition.

3. Nov 15, 2004

### astronut24

hello

well....yes, it's possible to have a cycle of length 2 in addition to the 5 cycle...but then the l.c.m becomes 10. so that rules out such a consideration!

4. Nov 15, 2004

### Galileo

Yes, so the only permutations in S7 of order 5 are the 5-cycles.

5. Nov 15, 2004

### matt grime

12345, 23451, 34512, 45123, and 51234 as different elements.

6. Nov 15, 2004

### astronut24

ya....so what further? that's a valid point you've raised...
so do you divide by 4?

7. Nov 15, 2004

### Galileo

You have ${7 \choose 5}$ ways to pick 5 elements from a set of 7.
There are 5! ways you can order 5 elements in a cycle.
For a given cycle of length 5, 5 orderings are give the same permutation.

8. Nov 16, 2004

### mansi

hey galileo

we left this a long while ago, right astronut?!

9. Nov 21, 2004

### mansi

i think the solution 504 is correct....
i don't see any fallacy in it.

10. Nov 21, 2004

### matt grime

The general solution is:

suppose t is a permutation of type

$$1^{m_1}2^{m_2}....r^{m_r}$$

then the order of the centralizer is

$$\prod_i i^{m_i}m_i!$$

in this case it is 1^2.5

so the centralizer's order is

2!.5

hence the conjugacy class has order

7!/10

which is indeed 504