# Orders of elements in Symmetric groups

• March 29th 2010, 08:49 PM
kimberu
Orders of elements in Symmetric groups
I'm reviewing for a midterm, and one of the questions is:

"What is the smallest such n so $S_n$ contains an element of order 30? Give reasons."

I don't know how to solve this. Clearly the order of the group must be at least 30, so n is greater than 4. Also 30 must divide the order of the group, by Lagrange's Theorem, but it does for every n greater than 4.

Thanks for any help!
• March 29th 2010, 08:54 PM
chiph588@
Quote:

Originally Posted by kimberu
I'm reviewing for a midterm, and one of the questions is:

"What is the smallest such n so $S_n$ contains an element of order 30? Give reasons."

I don't know how to solve this. Clearly the order of the group must be at least 30, so n is greater than 4. Also 30 must divide the order of the group, by Lagrange's Theorem, but it does for every n greater than 4.

Thanks for any help!

Here's a little hint:

Suppose $\sigma\in S_n$ has order $30$.
Then $30\mid |S_n| \implies 30\mid n!$.
• March 29th 2010, 09:01 PM
kimberu
Doesn't $30\mid n!$ for every n that is 5 or greater, though? Does that mean that n=5, and if so, why?
• March 29th 2010, 09:05 PM
chiph588@
Quote:

Originally Posted by kimberu
Doesn't $30\mid n!$ for every n that is 5 or greater, though? Does that mean that n=5, and if so, why?

Right, so start looking in $S_5$.

I suggest finding $\sigma\in S_5$ with order $5!=120$. Then $\sigma^4$ will be an element with order $30$.

Off the top of my head I can't think of such $\sigma$, but I imagine it's not to hard to find.
• March 30th 2010, 12:21 AM
Swlabr
Quote:

Originally Posted by chiph588@
Right, so start looking in $S_5$.

I suggest finding $\sigma\in S_5$ with order $5!=120$. Then $\sigma^4$ will be an element with order $30$.

Off the top of my head I can't think of such $\sigma$, but I imagine it's not to hard to find.

Pardon? $S_5$ is not cyclic! It is two-generated!

What you are supposed to figure out in this question is what the order of a permutation is. You know that a 2-cycle has order 2, a 3-cycle has order 3, etc. but you need to figure out what happens when you concatenate two or more cycles.

So, what is the order of $(12)(345)$? Why?

Similarly, what is the order of $(12)(34)$, and why?

What about $(1234)(56)$?

Your solution will involve splitting 30 down into prime factors.

-----

(I would like to point out at this stage that the element in $S_5$ with highest order is an element if the form $(12)(345)$).
• March 30th 2010, 12:34 AM
kimberu
Quote:

Originally Posted by Swlabr
So, what is the order of $(12)(345)$? Why?

Similarly, what is the order of $(12)(34)$, and why?

What about $(1234)(56)$?

Your solution will involve splitting 30 down into prime factors.

Those orders are 6, 2, and 4, I think, because those are the LCMs of the elements in the disjoint cycles. So since 30 = 2 x 3 x 5, there must be disjoint cycles of those amounts, and the smallest such case would be $S_{10}$?
• March 30th 2010, 12:40 AM
Swlabr
Quote:

Originally Posted by kimberu
Those orders are 6, 2, and 4, I think, because those are the LCMs of the elements in the disjoint cycles. So since 30 = 2 x 3 x 5, there must be disjoint cycles of those amounts, and the smallest such case would be $S_{10}$?

Exactly. The thing you should notice most is that it does not matter what numbers are in the disjoint cycle, it is what the disjoint cycle looks like. The same thing, curiously, happens for conjugacy classes,

Two permutations are conjugate in $S_n$ if and only if they have the same disjoint cycle shape.

For instance, $(12)(345)$, $(13)(245)$ and $(123)(45)$ are all conjugate in $S_n$ where $n \geq 5$.
• March 30th 2010, 02:39 PM
chiph588@
Quote:

Originally Posted by Swlabr
Pardon? $S_5$ is not cyclic! It is two-generated!

Whoops! How embarrassing! (Blush)