Permutations of S_36; Subgroup

[ESLASL1]

Homework Statement

Given Information: If sigma is a permutation of a set A, we say sigma moves "a" in set A iff sigma("a") is not equal to "a".

For the symmetric group S_36 of all permutations of 36 elements, let H be a subset of S_36 containing all permutations that move no more than for elements. Is H a subgroup of S_36? Prove.

Homework Equations

I understand that permutation is combination in a specific order but beyond that I am not sure what the problem is saying.

The Attempt at a Solution

I know I need to prove H is closed, the identity of S_36 (...move 4 elements...) is in H, and for every a in H a^-1 is also in H.

However, I am unclear as to what the original group "S_36 containing all permutations that move no more than for elements" is exactly. I am sure the "given information" line is suppose to clue me in but it means nothing to me. Nothing like like the feeling of stupid first thing in the morning.

Please, can anyone help?

 

[snipez90]

Intuitively it's easy to think of permutations as simply acting on objects. In this case, a permutation sigma "moves" an element a in A if it just takes a to some other object in b in A (so b =/= a). So in some sense this permutation acts "transitively" on A, actually moving each element. Note that the definition stipulates that sigma moves a in A if sigma(a) =/= a, which of course makes sense because we don't want sigma to fix each element.

Now return to the first step of your proof that the particular group of permutations specified in the problem is a subgroup. Specifically how many elements are moved by the identity? Would the problem change if in the problem statement "no more than four elements" was replaced by "exactly 4 elements", or by "a finite number of elements"?

 

[ESLASL1]

I'm sorry... I am really lost. Could the question be restated as: Is D_4 (group order 8) a subgroup of S_36?

 

[lanedance]

how about the 2 permutations (1234) and (5678), they both move 4 elements, now what is their product?

 

[jbunniii]

I'm sorry... I am really lost. Could the question be restated as: Is D_4 (group order 8) a subgroup of S_36?

No, because D_4 has 8 elements, whereas there are more than 8 elements of S_36 that move no more than four elements.

For example, there are

$$(36 \cdot 35 \cdot 34 \cdot 33) /4 = 353430$$

4-cycles in S_36, all of which move exactly 4 elements. These aren't even all of the elements that move exactly 4 elements, and there are many others that move 2 or 3.