Trying to simplify Nash's proof for the 2 player case (Symmetric game) 
Show that in any 2 player symmetric game there is always a symmetric
  Nash equilibrium where both players play the same strategy

I read the proof for n-player game in page 12 here and all the permutations made it really hard for me to read and understand..Does anyone have a proof or can help me simplify it for n=2?
Thanks
 A: In a non-degenerate game there is an indirect way through an argument provided by the Lemke Howson algorithm, namely that any non-degenerate bimatrix game (here I added finite to your assumptions, otherwise this seemingly obvious statement is not true: there are infinte/discontinuous symmetric games with only asymmetric Nash equilibria) has an odd number of equilibria. Hence, knowing that  


*

*there is an odd number of equilibria in any non-degenerate bimatrix game,

*for $x\neq y$, where $x,y$ are strategies, it is immediate to show that if $(x,y)$ is an equilibrium of a symmetric bimatrix game, then $(y,x)$ is an equilibrium as well


you can conclude that there is an even number of equilibria of the form $(x,y)$ with $x\neq y$ which imiplies that there must a $x^S$ such that $(x^S,x^S)$ is an equilibrium. 

Proof of point 2. Let $\alpha(x,y)$ denote the payoff function of player I, where $(x,y)\in X\times X$, where $X$ is the common strategy space. Let $\beta(x,y)$ denote the payoff function of player II. By symmetry 
$$\beta(x,y)=\alpha(y,x)\tag{sym}$$ Now, assume that $(x',y')$ is a Nash equilibrium of this bimatrix game, then $$\alpha(x',y')\ge\alpha(x,y'), \quad \forall x\in X\tag1$$ and $$\beta(x',y')\ge \beta(x',y), \quad \forall y \in X\tag2$$ Rewrite the conditions in view of the symmetric property $(\text{sym})$to obtain that $$\alpha(y',x')\ge \alpha(y,x'),\quad \forall y\in X\tag{2'}$$ and $$\beta(y',x')\ge \beta(y',x),\quad \forall x\in X\tag{1'}$$ Now, $(1')$ and $(2')$ together imply that $(y',x')$ is a Nash equilibrium of the simultaneous move game. 
