[Abstract Algebra] Permutations and shuffling cards

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Homework Help Overview

The problem involves a card-shuffling machine that rearranges a sequence of cards in a consistent manner. The original poster presents a scenario where hearts are shuffled twice, and the final order of the cards is given. The objective is to determine the order of the cards after the first shuffle based on the provided final arrangement.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the permutation resulting from the shuffling process and expresses uncertainty about how to determine the initial order of the cards. Some participants suggest considering the powers of the permutation to derive insights into the shuffling order.

Discussion Status

The discussion has progressed with participants exploring the properties of the permutation and its order. There is a recognition of the mathematical relationships involved, and some guidance has been offered regarding the implications of the permutation's cycle structure. However, the original poster's question about the first shuffle remains open for further exploration.

Contextual Notes

The problem is situated within the context of abstract algebra, specifically dealing with permutations and their properties. The original poster's inquiry is framed by the constraints of a homework assignment, which may limit the information available for discussion.

Klungo
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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


\

The Attempt at a Solution



We have a permutation
\alpha^2 = \left[ \begin{array}{cc}<br /> A &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp; 9 &amp; 10 &amp; J &amp; Q &amp; K \\<br /> 10 &amp; 9 &amp; Q &amp; 8 &amp; K &amp; 3 &amp; 4 &amp; A &amp; 5 &amp; J &amp; 6 &amp; 2 &amp; 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.
 
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Think about raising α2 to various powers.
 
haruspex said:
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).
 
Quite so - well done.
 
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?
 
bsoblick said:
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.
 

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