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KingGeorge

Group Theory Challenge Problem! Let \(S_n\) be the group of permutations on \(\{1,2,...,n\}\), and let two players play a game. Taking turns, the two players select elements one at a time from \(S_n\). Players may only select elements that have not already been selected. The game ends when the set of selected elements generate \(S_n\). The player who made the last move loses. Who wins the game? [HINT: Think of this problem in terms of the largest possible set of elements you can have so that you don't generate the whole group.] [HINT 2: What are the orders of the maximal subgroups?]

  • 10 months ago
  • 10 months ago

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  1. KingGeorge
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    This was a problem of the month offered by my undergraduate department where I was able to come up with the correct solution.

    • 10 months ago
  2. zzr0ck3r
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    I post an easier group theory problem and only @KingGeorge looks, @KingGeorge posts one and you all are like flies....

    • 10 months ago
  3. zzr0ck3r
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    im playing:)

    • 10 months ago
  4. zepdrix
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    If \(\large n\) is even, the player to select first would win. If \(\large n\) is odd, the player to select first would lose. Do I have the right idea?? D: No it's probably more math'y than that...

    • 10 months ago
  5. KingGeorge
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    Definitely more mathy than that. Although that was also one of my first ideas when given this question.

    • 10 months ago
  6. dan815
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    the player who made the 2nd last move

    • 10 months ago
  7. KingGeorge
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    Well, duh. But is that player the first or the second to have originally chosen?

    • 10 months ago
  8. swissgirl
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    Ummm can they remove more than one element at a time?

    • 10 months ago
  9. dan815
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    its the player that goes 2nd who wins

    • 10 months ago
  10. KingGeorge
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    Only one at a time @swissgirl Show me a proof @dan815

    • 10 months ago
  11. KingGeorge
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    Well, I'm off to bed. I'll leave you with the very vague hint, that while @zepdrix's idea is not the correct solution, you do need to keep even/odd in mind.

    • 10 months ago
  12. dan815
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    |dw:1370241063380:dw|

    • 10 months ago
  13. dan815
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    |dw:1370241133384:dw|

    • 10 months ago
  14. dan815
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    sooo the first one loses!

    • 10 months ago
  15. dan815
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    |dw:1370241271520:dw|

    • 10 months ago
  16. dan815
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    so all even but +1 elements from s1=1 will give u an odd number of elements in total

    • 10 months ago
  17. KingGeorge
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    I honestly have no idea what you're trying to do here.

    • 10 months ago
  18. dan815
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    well i m just kinda guessing what set of permutations are xD does that mean for each permutation say for 3, there are 3! elements in there

    • 10 months ago
  19. dan815
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    if it was like that in that case I said there are an even number of permutations in all but 1 element which is the 1! element so there will always be an odd number of elements in this set therefore the first one to Go, will always pick the first and last element

    • 10 months ago
  20. dan815
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    it cant be that easy so.. i probably dont understand what a group of permutations mean xD

    • 10 months ago
  21. KingGeorge
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    Hint: Think of this problem in terms of the largest possible set of elements you can have so that you don't generate the whole group.

    • 10 months ago
  22. KingGeorge
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    Hint 2: What are the orders of the maximal subgroups?

    • 10 months ago
  23. alexandercpark
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    Would the size of s_n be nCn?

    • 9 months ago
  24. KingGeorge
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    The size of \(S_n\) is \(n!\).

    • 9 months ago
  25. alexandercpark
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    forgive my lack of talent in the equation editor :( because the size of s_n is n!, the ONLY time that the size of s_n is odd is when s_n is the emty set, or the set {1} any set beyond that would be even, because you have (integers) * 2. so, the winner depends on the origanl size of s_n, if n = 0 or 1, the first player to go will lose. if n > 1, the second player to go will lose.

    • 9 months ago
  26. KingGeorge
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    There are some additional considerations you have to take in mind since this is a group. If you aren't familiar with some group theory, this problem is going to give you a tough time.

    • 9 months ago
  27. domu
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    The set of n-1 elements \[\left\{ (j,j+1) \left| \right| 1 \le j < n \right\}\] form a generating set. Thus, for n odd, the first player losses and for n even the second player losses. I believe this is correct but I still need to show that a set of n-2 elements does not generate the whole group. Thought? Counterexamples?

    • 9 months ago
  28. KingGeorge
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    That is not correct. For example, if n=3, then I will choose \((123)\). Then, in order not to lose, second player must choose either \(e\) or \((132)\). I then choose the one second player didn't choose. Then whatever second player chooses next, he loses. So I win as first player.

    • 9 months ago
  29. KingGeorge
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    Since this seems to have died, I'll be posting the solution here. For \(n=2,3\) player 1 wins. For any other \(n\ge1\), player 2 wins. It should be straightforward to see why this is the case for \(n=1,2,3\). Now suppose \(n\ge4\). First, I will prove that any maximal subgroup of \(S_n\) has even order. Note that any subgroup \(H\) of \(S_n\) that has odd order must have only elements of odd order as well. Let \(w\) be such an element. Then consider the unique cycle-decomposition of \(w\). Since \(|w|\) is odd, all the cycles must also have odd order. But if a cycle has odd order, then it must be contained in the alternating subgroup \(A_n\). Thus, \(w\in A_n\), and \(H\subsetneq A_n\). So \(H\) is not maximal. Thus, any maximal subgroup has even order. Now suppose \(2k\) elements have been chosen from \(S_n\) for some \(k\in\mathbb{Z}_{\ge0}\) so it is player 1's turn. We have two cases. Case 1: Player 1 chooses so that the game ends. Then player 2 wins. Case 2: Player 1 chooses so that the game does not end. Then \(2k+1\) elements have been chosen. But since every maximal subgroup has even order, this set of elements must be properly contained in some maximal subgroup. So there is at least one more element \(g\) in that subgroup that has not been chosen. Player 2 may choose \(g\), and thus does not end the game. We now have chosen \(2(k+1)\) elements. Since there are a finite number of elements, we are done by induction.

    • 8 months ago
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