Raise the following permutations to as many powers as it takes to get to the identity permutation. (That is, for each permutation, write it, its square, the permutation you get if you iterate it three times, etc. until you get to the identity. Each permutation or power of a permutation in your answer should be written as a product of DISJOINT cycles.)
(1 2 3 4)(4 5 6) in S_7

Question

Raise the following permutations to as many powers as it takes to get to the identity permutation.  (That is, for each permutation, write it, its square, the permutation you get if you iterate it three times, etc. until you get to the identity.  Each permutation or power of a permutation in your answer should be written as a product of DISJOINT cycles.)
(1 2 3 4)(4 5 6) in S_7

anonymous · Answer

If we call the product $\alpha$, then it looks like $\alpha^6$ gives the identity $(1)(2)(3)(4)(5)(6)(7)$.

anonymous · Answer

Apparently there is a theorem for this that says the order of a permutation (the number of the exponent needed to get $\alpha^n$ to be equivalent to the identity) is the least common multiple of the lengths of each cycle. In this case, you have two cycles of length 4 and 3, which would make 12 the necessary number of iterations, but that's not what my work suggests...

anonymous · Answer

I believe it has something to do with the fact that 4 is part of both cycles. I've only just read up on the subject, and the examples only show permutations of non-repeats.

anonymous · Answer

I haven't heard of that theorem, but I'll look into it. Do you know how to get to the identity function from cycle notation?

anonymous · Answer

Do you know how to find the product of two cycles?

anonymous · Answer

I had just learned the concept 30 min ago, but I think I'd be able to get at least the computational aspect across. Not so sure about the theoretical.

anonymous · Answer

We briefly went over that, but yes I believe I know how to find the product. Apologies for the slow response

anonymous · Answer

Well I suppose it wouldn't hurt to show you what I did. Given $\alpha=(1,2,3,4)(4,5,6)\color{gray}{(7)}$ (I'll omit the (7) from here on), you have $\alpha^2=(1,2,3,4)(4,5,6)(1,2,3,4)(4,5,6)$, which gives
$$\begin{matrix}1\xrightarrow{(1,2,3,4)}2\xrightarrow{(4,5,6)}2\xrightarrow{(1,2,3,4)}3\xrightarrow{(4,5,6)}3&\rceil\
3\xrightarrow{(1,2,3,4)}4\xrightarrow{(4,5,6)}5\xrightarrow{(1,2,3,4)}5\xrightarrow{(4,5,6)}6&|&\rightarrow&(1,3,6)\
6\xrightarrow{(1,2,3,4)}6\xrightarrow{(4,5,6)}4\xrightarrow{(1,2,3,4)}1\xrightarrow{(4,5,6)}1&\rfloor \
2\xrightarrow{(1,2,3,4)}3\xrightarrow{(4,5,6)}3\xrightarrow{(1,2,3,4)}4\xrightarrow{(4,5,6)}5&\rceil\
5\xrightarrow{(1,2,3,4)}5\xrightarrow{(4,5,6)}6\xrightarrow{(1,2,3,4)}6\xrightarrow{(4,5,6)}4&|&\rightarrow&(2,5,4)\
4\xrightarrow{(1,2,3,4)}1\xrightarrow{(4,5,6)}1\xrightarrow{(1,2,3,4)}2\xrightarrow{(4,5,6)}2&\rfloor
\end{matrix}$$
And so $\alpha^2=(1,3,6)(2,5,4)$. Rinse and repeat... until you find that $\alpha^6$ gives the identity.

anonymous · Answer

That looks similar to what I've seen, I'm having trouble seeing how you get the product from permutations. For the first part...
1->2->2->3->3 
3->4->5->6
6->6
Okay I see what you're doing.

anonymous · Answer

How do you know once you've reached the identity function?

anonymous · Answer

You'll know when each number gets mapped back to itself in one line, meaning you should see something like this:
$$1\xrightarrow{\text{1st cycle}}\begin{pmatrix}\text{whatever 1 gets}\\text{mapped to in the}\\text{1st cycle}\end{pmatrix}\xrightarrow{\text{2nd cycle}}\cdots\xrightarrow{n\text{th cycle}}1$$

anonymous · Answer

@SithsAndGiggles  Is there  a difference between this permutation in s_7 as oppose to if it were in s_8?