- 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?]

- schrodinger

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- KingGeorge

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

- zzr0ck3r

I post an easier group theory problem and only @KingGeorge looks, @KingGeorge posts one and you all are like flies....

- zzr0ck3r

im playing:)

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## More answers

- zepdrix

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...

- KingGeorge

Definitely more mathy than that. Although that was also one of my first ideas when given this question.

- dan815

the player who made the 2nd last move

- KingGeorge

Well, duh. But is that player the first or the second to have originally chosen?

- swissgirl

Ummm can they remove more than one element at a time?

- dan815

its the player that goes 2nd who wins

- KingGeorge

Only one at a time @swissgirl
Show me a proof @dan815

- KingGeorge

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.

- dan815

|dw:1370241063380:dw|

- dan815

|dw:1370241133384:dw|

- dan815

sooo the first one loses!

- dan815

|dw:1370241271520:dw|

- dan815

so all even but +1 elements from s1=1 will give u an odd number of elements in total

- KingGeorge

I honestly have no idea what you're trying to do here.

- dan815

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

- dan815

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

- dan815

it cant be that easy so.. i probably dont understand what a group of permutations mean xD

- KingGeorge

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.

- KingGeorge

Hint 2: What are the orders of the maximal subgroups?

- anonymous

Would the size of s_n be nCn?

- KingGeorge

The size of \(S_n\) is \(n!\).

- anonymous

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.

- KingGeorge

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.

- anonymous

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?

- KingGeorge

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.

- KingGeorge

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.

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