# Orders of elements in Symmetric groups

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

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

Right, so start looking in $\displaystyle S_5$.

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

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

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

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

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

Pardon? $\displaystyle S_5$ 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 $\displaystyle (12)(345)$? Why?

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

What about $\displaystyle (1234)(56)$?

Your solution will involve splitting 30 down into prime factors.

-----

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

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

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

What about $\displaystyle (1234)(56)$?

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 $\displaystyle S_{10}$?
• Mar 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 $\displaystyle S_{10}$?

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 $\displaystyle S_n$ if and only if they have the same disjoint cycle shape.

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

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

Whoops! How embarrassing! (Blush)