Let $\alpha = (1 6 2) (4 5)$ and $\beta = (1 2 3 6)$ be permutations in $S_ {6}$. Calculate the product $\beta \alpha$. Answer with a product of disjunked cycles.
I don't know how to do it, but maybe something like this;
Step 1: I think that if we look at $\alpha(1)$, we see that in $\beta$ that that answers to 2. And 2 in $\alpha$ is equal to 1, so that is our first cycle, (1 2).
Step two: And then we look at $\alpha(2)$ and that is in $\beta$=1 and that is in $\alpha$ = 6.. So that's another cycle?
Step three: We're no looking at $\alpha(3)$ with is in $\beta$=6. So there's another cycle (3 6) and we don't have a 3 in $\alpha$ so this is another cycle..
Step four: $\alpha(4), \alpha(5)$ dosen't have any 'friends' in $\beta$ so this is the other cycle (4)(5).
Therefore, the answer should be (1 2)(2 6)(3)(4)(5)
hhmmm.. But it's wrong. Someone help me out? :)
So we have: $α=(162)(45)$
$β=(1236)$
$\alpha \beta =(1)(23)(45)(6)=(23)(45)$
You do this: you look at where 1 get map to by $\beta$, it gets map to 2, then you look at where 2 gets map to by $\alpha$, it gets map to 1. So 1 goes to 1 by $\alpha\beta$. Then you look at where 2 gets map to by $\beta$, it gets map to 3 and then you look at where 3 gets map to by $\alpha$, it is being fixed. so 2 gets map to 3 by $\alpha\beta$. Then you look at where 3 gets map to by $\beta$, it gets map to 6, and you look at where 6 gets map to by $\alpha$, it gets map to 2. So 3 gets map to 2 by $\alpha\beta$ and you close the cycle. You keep doing this until you know where every numbers gets map to to get a disjoint union of cycles. And then you can remove cycles of length 1 because they are redundant as you can just say numbers not present in your cycle decomposition gets map to themselves.