Developing a strategy to win a game of picking elements from $S_n$
$begingroup$
Given a integer $n>1$, Let $S_n$ be the group of permutations of the numbers $1,2,dots n$. Two players, $A$ and $B$, play the following game. Taking turns, they select elements(one element at a time) from the group $S_n$. It is forbidden to select an element that had been already selected. The game ends when the selected elements generate the whole group $S_n$. The player who made the last move loses. The first move is made by $A$. Which player has a winning strategy?
My attempt involves finding
What elements of $S_n$ can generate $S_n$?
I know that $(123 dots n)$ and $(12)$ can generate $S_n$.
But we are supposed to look for all other such set of elements which can generate $S_n$.
How do I solve this?
group-theory discrete-mathematics combinatorial-game-theory
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
$endgroup$
add a comment |
$begingroup$
Given a integer $n>1$, Let $S_n$ be the group of permutations of the numbers $1,2,dots n$. Two players, $A$ and $B$, play the following game. Taking turns, they select elements(one element at a time) from the group $S_n$. It is forbidden to select an element that had been already selected. The game ends when the selected elements generate the whole group $S_n$. The player who made the last move loses. The first move is made by $A$. Which player has a winning strategy?
My attempt involves finding
What elements of $S_n$ can generate $S_n$?
I know that $(123 dots n)$ and $(12)$ can generate $S_n$.
But we are supposed to look for all other such set of elements which can generate $S_n$.
How do I solve this?
group-theory discrete-mathematics combinatorial-game-theory
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
$endgroup$
$begingroup$
"all other" sets might be a bit ambitious.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 4:56
$begingroup$
The probability that two random elements of $S_n$ generate $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Dec 23 '18 at 10:41
$begingroup$
Please ask one question at a time.
$endgroup$
– Shaun
Dec 23 '18 at 16:11
$begingroup$
@shaun The actual question is "Second Question". I thought that the first question is necessary to solve the actual question. My attempt involves solving first question to solve second question
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 17:04
1
$begingroup$
Okay. I suggest that you edit the question, then, to make that clear. It might be closed as too broad otherwise, since it's easy to overlook such detail.
$endgroup$
– Shaun
Dec 23 '18 at 17:06
add a comment |
$begingroup$
Given a integer $n>1$, Let $S_n$ be the group of permutations of the numbers $1,2,dots n$. Two players, $A$ and $B$, play the following game. Taking turns, they select elements(one element at a time) from the group $S_n$. It is forbidden to select an element that had been already selected. The game ends when the selected elements generate the whole group $S_n$. The player who made the last move loses. The first move is made by $A$. Which player has a winning strategy?
My attempt involves finding
What elements of $S_n$ can generate $S_n$?
I know that $(123 dots n)$ and $(12)$ can generate $S_n$.
But we are supposed to look for all other such set of elements which can generate $S_n$.
How do I solve this?
group-theory discrete-mathematics combinatorial-game-theory
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
$endgroup$
Given a integer $n>1$, Let $S_n$ be the group of permutations of the numbers $1,2,dots n$. Two players, $A$ and $B$, play the following game. Taking turns, they select elements(one element at a time) from the group $S_n$. It is forbidden to select an element that had been already selected. The game ends when the selected elements generate the whole group $S_n$. The player who made the last move loses. The first move is made by $A$. Which player has a winning strategy?
My attempt involves finding
What elements of $S_n$ can generate $S_n$?
I know that $(123 dots n)$ and $(12)$ can generate $S_n$.
But we are supposed to look for all other such set of elements which can generate $S_n$.
How do I solve this?
group-theory discrete-mathematics combinatorial-game-theory
group-theory discrete-mathematics combinatorial-game-theory
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
edited Dec 23 '18 at 17:09
Rakesh Bhatt
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
asked Dec 23 '18 at 4:40
Rakesh BhattRakesh Bhatt
967214
967214
$begingroup$
"all other" sets might be a bit ambitious.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 4:56
$begingroup$
The probability that two random elements of $S_n$ generate $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Dec 23 '18 at 10:41
$begingroup$
Please ask one question at a time.
$endgroup$
– Shaun
Dec 23 '18 at 16:11
$begingroup$
@shaun The actual question is "Second Question". I thought that the first question is necessary to solve the actual question. My attempt involves solving first question to solve second question
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 17:04
1
$begingroup$
Okay. I suggest that you edit the question, then, to make that clear. It might be closed as too broad otherwise, since it's easy to overlook such detail.
$endgroup$
– Shaun
Dec 23 '18 at 17:06
add a comment |
$begingroup$
"all other" sets might be a bit ambitious.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 4:56
$begingroup$
The probability that two random elements of $S_n$ generate $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Dec 23 '18 at 10:41
$begingroup$
Please ask one question at a time.
$endgroup$
– Shaun
Dec 23 '18 at 16:11
$begingroup$
@shaun The actual question is "Second Question". I thought that the first question is necessary to solve the actual question. My attempt involves solving first question to solve second question
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 17:04
1
$begingroup$
Okay. I suggest that you edit the question, then, to make that clear. It might be closed as too broad otherwise, since it's easy to overlook such detail.
$endgroup$
– Shaun
Dec 23 '18 at 17:06
$begingroup$
"all other" sets might be a bit ambitious.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 4:56
$begingroup$
"all other" sets might be a bit ambitious.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 4:56
$begingroup$
The probability that two random elements of $S_n$ generate $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Dec 23 '18 at 10:41
$begingroup$
The probability that two random elements of $S_n$ generate $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Dec 23 '18 at 10:41
$begingroup$
Please ask one question at a time.
$endgroup$
– Shaun
Dec 23 '18 at 16:11
$begingroup$
Please ask one question at a time.
$endgroup$
– Shaun
Dec 23 '18 at 16:11
$begingroup$
@shaun The actual question is "Second Question". I thought that the first question is necessary to solve the actual question. My attempt involves solving first question to solve second question
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 17:04
$begingroup$
@shaun The actual question is "Second Question". I thought that the first question is necessary to solve the actual question. My attempt involves solving first question to solve second question
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 17:04
1
1
$begingroup$
Okay. I suggest that you edit the question, then, to make that clear. It might be closed as too broad otherwise, since it's easy to overlook such detail.
$endgroup$
– Shaun
Dec 23 '18 at 17:06
$begingroup$
Okay. I suggest that you edit the question, then, to make that clear. It might be closed as too broad otherwise, since it's easy to overlook such detail.
$endgroup$
– Shaun
Dec 23 '18 at 17:06
add a comment |
1 Answer
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$begingroup$
I guess for $n=1$, A must choose the identity, which generates $S_n$, so B wins.
For $n=2$ A wins by choosing the identity, and for $n=3$ A wins by choosing a $3$-cycle, such as $(1,2,3)$.
For $n ge 4$, all maximal subgroups of $S_n$ have even order, and the subgroup generated by the elements chosen so far will eventually be maximal, so B wins.
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
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add a comment |
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$begingroup$
I guess for $n=1$, A must choose the identity, which generates $S_n$, so B wins.
For $n=2$ A wins by choosing the identity, and for $n=3$ A wins by choosing a $3$-cycle, such as $(1,2,3)$.
For $n ge 4$, all maximal subgroups of $S_n$ have even order, and the subgroup generated by the elements chosen so far will eventually be maximal, so B wins.
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
$endgroup$
add a comment |
$begingroup$
I guess for $n=1$, A must choose the identity, which generates $S_n$, so B wins.
For $n=2$ A wins by choosing the identity, and for $n=3$ A wins by choosing a $3$-cycle, such as $(1,2,3)$.
For $n ge 4$, all maximal subgroups of $S_n$ have even order, and the subgroup generated by the elements chosen so far will eventually be maximal, so B wins.
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
$endgroup$
add a comment |
$begingroup$
I guess for $n=1$, A must choose the identity, which generates $S_n$, so B wins.
For $n=2$ A wins by choosing the identity, and for $n=3$ A wins by choosing a $3$-cycle, such as $(1,2,3)$.
For $n ge 4$, all maximal subgroups of $S_n$ have even order, and the subgroup generated by the elements chosen so far will eventually be maximal, so B wins.
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
$endgroup$
I guess for $n=1$, A must choose the identity, which generates $S_n$, so B wins.
For $n=2$ A wins by choosing the identity, and for $n=3$ A wins by choosing a $3$-cycle, such as $(1,2,3)$.
For $n ge 4$, all maximal subgroups of $S_n$ have even order, and the subgroup generated by the elements chosen so far will eventually be maximal, so B wins.
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
edited Dec 23 '18 at 18:18
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
answered Dec 23 '18 at 17:26
Derek HoltDerek Holt
54.7k53574
54.7k53574
add a comment |
add a comment |
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$begingroup$
"all other" sets might be a bit ambitious.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 4:56
$begingroup$
The probability that two random elements of $S_n$ generate $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Dec 23 '18 at 10:41
$begingroup$
Please ask one question at a time.
$endgroup$
– Shaun
Dec 23 '18 at 16:11
$begingroup$
@shaun The actual question is "Second Question". I thought that the first question is necessary to solve the actual question. My attempt involves solving first question to solve second question
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 17:04
1
$begingroup$
Okay. I suggest that you edit the question, then, to make that clear. It might be closed as too broad otherwise, since it's easy to overlook such detail.
$endgroup$
– Shaun
Dec 23 '18 at 17:06