
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 $\displaystyle \sigma=$(1 2 3 4 5 6) in $\displaystyle S_6$. Show that $\displaystyle G=\{\epsilon,\sigma, \sigma^2,\sigma^3,\sigma^4,\sigma^5\}$ is a group using the operation of $\displaystyle S_6$. Is G abelian? How many elements $\displaystyle \tau$ of G satisfy $\displaystyle \tau^2=\epsilon$ and $\displaystyle \tau^3=\epsilon$?
I said that the operation of $\displaystyle S_6$ is closed and associative, that every element in $\displaystyle S_6$ has an inverse, and that $\displaystyle \epsilon=1_G$. 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?

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.

(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?