Game involving taking exponent of a group until it becomes trivial.
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Let $G neq {1}$ be a finite group. Two players I $&$ II, that know the group $G$ are playing the following game: Player I chooses a prime $p_1$ and then the players consider the group $G(p_1)):= G^{p_1}$. Player II chooses a prime $q_1$ and they consider the group $G(p_1,q_1):=(G^{P_1})^{q_1}$. Player I, then chooses a prime $p_2$ and they consider $G(p_1,q_1,p_2)=((G^{P_1})^{q_1})^{p_2}$ and so on. The first player to reach the trivial group wins. That is, if for some $p_i$, $G(P_1,...,q_{i-1},p_i)={1}$ but $G(p_1,...,q_{i-1}) neq {1}$, player I had won. Similarly for player II.
a) Prove that player II does not have a strategy that guarantees him a win no matter what the group $G$ is.
b) Suppose now that $G$ is abelian and let us also add the constraint that at every stage the players have to choose a prime that divides the order of the group at that stage. Provide a necessary and sufficient condition on $G$ for player I to have a winning strategy.
abstract-algebra group-theory finite-groups combinatorial-game-theory
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
asked Nov 18 at 2:27
mathnoob
1,133116
add a comment |
up vote
2
down vote
favorite
Let $G neq {1}$ be a finite group. Two players I $&$ II, that know the group $G$ are playing the following game: Player I chooses a prime $p_1$ and then the players consider the group $G(p_1)):= G^{p_1}$. Player II chooses a prime $q_1$ and they consider the group $G(p_1,q_1):=(G^{P_1})^{q_1}$. Player I, then chooses a prime $p_2$ and they consider $G(p_1,q_1,p_2)=((G^{P_1})^{q_1})^{p_2}$ and so on. The first player to reach the trivial group wins. That is, if for some $p_i$, $G(P_1,...,q_{i-1},p_i)={1}$ but $G(p_1,...,q_{i-1}) neq {1}$, player I had won. Similarly for player II.
a) Prove that player II does not have a strategy that guarantees him a win no matter what the group $G$ is.
b) Suppose now that $G$ is abelian and let us also add the constraint that at every stage the players have to choose a prime that divides the order of the group at that stage. Provide a necessary and sufficient condition on $G$ for player I to have a winning strategy.
abstract-algebra group-theory finite-groups combinatorial-game-theory
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
asked Nov 18 at 2:27
mathnoob
1,133116
I presume by $G^p$ you mean ${ g^p : g in G }$. This isn't a subgroup in general unless $G$ is abelian. Strictly speaking that doesn't affect the definition of the game, though.
– Qiaochu Yuan
Nov 18 at 3:44
yes, That is correct.
– mathnoob
Nov 18 at 11:39
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $G neq {1}$ be a finite group. Two players I $&$ II, that know the group $G$ are playing the following game: Player I chooses a prime $p_1$ and then the players consider the group $G(p_1)):= G^{p_1}$. Player II chooses a prime $q_1$ and they consider the group $G(p_1,q_1):=(G^{P_1})^{q_1}$. Player I, then chooses a prime $p_2$ and they consider $G(p_1,q_1,p_2)=((G^{P_1})^{q_1})^{p_2}$ and so on. The first player to reach the trivial group wins. That is, if for some $p_i$, $G(P_1,...,q_{i-1},p_i)={1}$ but $G(p_1,...,q_{i-1}) neq {1}$, player I had won. Similarly for player II.
a) Prove that player II does not have a strategy that guarantees him a win no matter what the group $G$ is.
b) Suppose now that $G$ is abelian and let us also add the constraint that at every stage the players have to choose a prime that divides the order of the group at that stage. Provide a necessary and sufficient condition on $G$ for player I to have a winning strategy.
abstract-algebra group-theory finite-groups combinatorial-game-theory
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
asked Nov 18 at 2:27
mathnoob
1,133116
Let $G neq {1}$ be a finite group. Two players I $&$ II, that know the group $G$ are playing the following game: Player I chooses a prime $p_1$ and then the players consider the group $G(p_1)):= G^{p_1}$. Player II chooses a prime $q_1$ and they consider the group $G(p_1,q_1):=(G^{P_1})^{q_1}$. Player I, then chooses a prime $p_2$ and they consider $G(p_1,q_1,p_2)=((G^{P_1})^{q_1})^{p_2}$ and so on. The first player to reach the trivial group wins. That is, if for some $p_i$, $G(P_1,...,q_{i-1},p_i)={1}$ but $G(p_1,...,q_{i-1}) neq {1}$, player I had won. Similarly for player II.
a) Prove that player II does not have a strategy that guarantees him a win no matter what the group $G$ is.
b) Suppose now that $G$ is abelian and let us also add the constraint that at every stage the players have to choose a prime that divides the order of the group at that stage. Provide a necessary and sufficient condition on $G$ for player I to have a winning strategy.
abstract-algebra group-theory finite-groups combinatorial-game-theory
abstract-algebra group-theory finite-groups combinatorial-game-theory
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
asked Nov 18 at 2:27
mathnoob
1,133116
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
asked Nov 18 at 2:27
mathnoob
1,133116
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
edited Nov 18 at 3:23
Eric Wofsey
176k12202326
176k12202326
asked Nov 18 at 2:27
mathnoob
1,133116
asked Nov 18 at 2:27
mathnoob
1,133116
asked Nov 18 at 2:27
mathnoob
1,133116
1,133116
I presume by $G^p$ you mean ${ g^p : g in G }$. This isn't a subgroup in general unless $G$ is abelian. Strictly speaking that doesn't affect the definition of the game, though.
– Qiaochu Yuan
Nov 18 at 3:44
yes, That is correct.
– mathnoob
Nov 18 at 11:39
add a comment |
I presume by $G^p$ you mean ${ g^p : g in G }$. This isn't a subgroup in general unless $G$ is abelian. Strictly speaking that doesn't affect the definition of the game, though.
– Qiaochu Yuan
Nov 18 at 3:44
yes, That is correct.
– mathnoob
Nov 18 at 11:39
I presume by $G^p$ you mean ${ g^p : g in G }$. This isn't a subgroup in general unless $G$ is abelian. Strictly speaking that doesn't affect the definition of the game, though.
– Qiaochu Yuan
Nov 18 at 3:44
I presume by $G^p$ you mean ${ g^p : g in G }$. This isn't a subgroup in general unless $G$ is abelian. Strictly speaking that doesn't affect the definition of the game, though.
– Qiaochu Yuan
Nov 18 at 3:44
yes, That is correct.
– mathnoob
Nov 18 at 11:39
yes, That is correct.
– mathnoob
Nov 18 at 11:39
add a comment |
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I presume by $G^p$ you mean ${ g^p : g in G }$. This isn't a subgroup in general unless $G$ is abelian. Strictly speaking that doesn't affect the definition of the game, though.
– Qiaochu Yuan
Nov 18 at 3:44
yes, That is correct.
– mathnoob
Nov 18 at 11:39