# Permutations in S_6

• Apr 4th 2011, 06:10 PM
manygrams
Permutations in S_6
Hey there,

I'm having a bit of trouble with a practise question from my abstract algebra class.

The question is:

Let in $\sigma=$(1 2 3 4 5 6) in $S_6$. Show that $G=\{\epsilon,\sigma, \sigma^2,\sigma^3,\sigma^4,\sigma^5\}$ is a group using the operation of $S_6$. Is G abelian? How many elements $\tau$ of G satisfy $\tau^2=\epsilon$ and $\tau^3=\epsilon$?

I said that the operation of $S_6$ is closed and associative, that every element in $S_6$ has an inverse, and that $\epsilon=1_G$. I think this is enough to show that G is a group. However, I'm not sure what to do for the abelian or last part. I know that multiplication of disjoint cycles is abelian, but the elements in G that aren't the identity aren't disjoint. I'm kind of lost.

Any hints?
• Apr 4th 2011, 08:53 PM
mathstew
Be careful. You need to show that G is closed, not S6 (obviously S6 is closed since it is a group). The operation on S6 is function composition. To show that G is a group is trivial. Since all you have is powers of sigma (I will now denote by s) (s^n)(s^m) = (s^r) where r = m+n (mod 6). Thus it is clear that every element has an inverse since (s)(s^5)=s^6=e, (s^2)(s^4)=e, and (s^3)(s^3)=e.

Also G is abelian since it doesnt matter which permutation you take first, they are just powers of sigma. The last part is easy as well, clearly s^3 is the only element of order 2. The elements of order 3 are s^2 and s^4.
• Apr 5th 2011, 01:46 AM
Deveno
(1 2 3 4 5 6) IS a disjoint cycle, of length 6.

to be fair, showing closure of G also involves proving σ^6 = e (so that the powers of σ listed are all there are).

after that is shown, the rest is easy, since G is just <σ>. not only is G abelian, it is *cyclic*. it is isomorphic to Z6.

here is the isomorphism φ:Z6-->G, φ(k) = σ^k. what are the elements of order 2 in Z6? what are the elements of order 3?