Game involving taking exponent of a group until it becomes trivial.

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.

• 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 '18 at 3:44
• yes, That is correct. – mathpadawan Nov 18 '18 at 11:39

(a): There might be a strategy-stealing argument here, but I'm not convinced that I understand the question completely. It seems like the game does not terminate, since either player can always just "stall" by choosing a prime $$p$$ which is $$1\bmod{|G|}$$, and by Lagrange's Theorem, $$G^p=G$$. Even if we impose the restriction that the primes must be distinct, Dirichlet's Theorem permits infinite stalling.
(b): Count the number of primes which divide $$G$$. If the number is odd, player I wins; if the number is even, player II wins. This doesn't seem to have anything to do with strategy: the outcomes are guaranteed no matter what moves the players make!