# Orders of elements

• Jan 9th 2009, 10:45 AM
daaavo
Orders of elements
Hi,

I was just wondering if you could give me some help to get me started on some more complex problems than the one below. If you could answer the question below and show me how it's to be done it would be much appreciated,

List the elements of S3 and their orders
• Jan 9th 2009, 11:09 AM
bulls6x
I'm guessing you mean \$\displaystyle S_3 \$ Well the elements are just all the permutations of \$\displaystyle \{1,2,3\} \$ which are:

\$\displaystyle (1), (1 2), (1 3), (2 3), (1 2 3), (1 3 2) \$

Now just compute the order of each element. For example,

\$\displaystyle \mid(1 2 3)\mid = 3 \$ since \$\displaystyle (1 2 3)(1 2 3)(1 2 3)= (1 2 3)(1 3 2) = (1) \$
• Jan 9th 2009, 11:15 AM
HallsofIvy
Quote:

Originally Posted by daaavo
Hi,

I was just wondering if you could give me some help to get me started on some more complex problems than the one below. If you could answer the question below and show me how it's to be done it would be much appreciated,

List the elements of S3 and their orders

Well, the first thing you should do is check on the definition of S3!

It is group of permutations of 3 objects with composition as the operation.
If the 3 objects are a, b, and c then the 6 ways of "permuting" them are
abc, acb, bac, bca, cab, and cba.

Now check the definition of order: a member, a, of a group is said to have order n if n is the smallest integer such that an= e, the group identity.

So: for each of the member of S3 above, check that. (abc) is, of course, the identity: its "order" is 1. (acb) where we just swap b and c is also easy: doing it twice just swaps them again: (acb)(acb)= (abc), the identity, so the order of (acb) is 2. Similarly (bac) just swaps a and b so its order is also 2. (cba) swaps a and c so its order is 2. (bca) rotates everything to the left so doing it twice results in (cab) and one more time (abc): the order of (bca) is 3 and the same is true of (cab) which rotates to the right: \$\displaystyle (cab)^2= (bca)\$ and \$\displaystyle (cab)^3= abc\$.

(I see that bull6x has said the same thing using a slightly different notation for the permutations.)