My problem is to show that any free group $F_{n}$ has a normal subgroup of index 2. I know that any subgroup of index 2 is normal. But how do I find a subgroup of index 2?
The subgroup needs to have 2 cosets. My first guess is to construct a subgroup $H<G$ as $H = <x_{1}^{2}, x_{2}^{2}, … , x_{n}^{2} >$ but this wouldn't be correct because $x_{1}H \ne x_{2}H \ne H$.
What is a way to construct such a subgroup?
Best Answer
Use the universal property of free groups: define a set theoretical function
$$f:\{x_1,...,x_n\}\to C_2=\langle c\rangle\,\,,\,\,f(x_1):=c\,\,,f(x_i)=1\,\,\,\forall\,i=2,3,...,n$$
Then there exists a unique group homomorphism
$$\phi: F_n\to C_2\,\;\; s.t. \;\;\,\phi(x_i)=f(x_i)\,\,\,\forall\,i=1,2,....,n\,$$
Well, $\,\ker\phi\,$ is your guy...