[Abstract Algebra] Permutations and shuffling cards

[Klungo]

It's been a while since I've posted. This is a problem I had for a homework assignment a few weeks ago but I completely figure out. Any help appreciated.

Homework Statement

"A card-shuffling machine always rearranges cards in the same way relative to the order
in which they were given to it. All of the hearts arranged in order
from ace to king were put into the machine, and then the shuffled
cards were put into the machine again to be shuffled. If the cards
emerged in the order 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7, in what
order were the cards after the first shuffle?"

Homework Equations

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The Attempt at a Solution

We have a permutation
$\alpha^2 = \left[ \begin{array}{cc} A & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & J & Q & K \\ 10 & 9 & Q & 8 & K & 3 & 4 & A & 5 & J & 6 & 2 & 7 \end{array} \right]$ which can be written as a single 13-cycle $\alpha^2 = (A,10,J,6,3,Q,2,9,5,K,7,4,8)$.

The goal is to find out how where each card is mapped to under $\alpha$.

And I'm not sure how to go about this.

 

[haruspex]

Think about raising α² to various powers.

 

[Klungo]

Think about raising α² to various powers.

From what I recalled on my first attempt.

We know that $(\alpha^2)^{13} = \alpha^{26} = e$ since $\alpha^2$ is a 13-single cycle. So, $|\alpha^{2}| \mbox{ divides } 13$.

That is, $|\alpha^2| = 1 \mbox{ or } 13$. Clearly, $|\alpha^2| \neq 1$ since $\alpha^2 \neq e$. Thus, $|\alpha^2| = 13$.

Now that I thought about it.

Using the formula $|g| = |g^k|gcd(k,|g|)$, we see that $|\alpha| = |\alpha^{2}| gcd(2,|\alpha|) = 13 gcd(2,|\alpha|) = 26 \mbox{ or } 13$.

Since we're "working" with the symmetric group $S_{13}$ and $26 = 2 \times 13$, we know that no element (by product of cycles) in $S_{13}$ has order 26. So, $|\alpha|=13$.

Hence, $\alpha = \alpha^{13} \alpha = \alpha^{14} = (\alpha^{2})^{7}$. Therefore, starting with A, we find $\alpha = (A,9,10,5,J,K,6,7,3,4,Q,8,2)$.

 

[haruspex]

Quite so - well done.

 

[bsoblick]

So does this imply that the order of the cards after the first shuffle were 9 A 4 Q J 7 3 2 10 5 K 8 6?

 

[haruspex]

So does this imply that the order of the cards after the first shuffle were 9 A 4 Q J 7 3 2 10 5 K 8 6?

Yes.