Free group n contains subgroup of index 2 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?    

For any subgroup $H$ of $G$ and elements $a$ and $b$ of $G$ the following statements hold.


*

*If $a \in H$ and $b \in H$, then $ab \in H$

*If $a \in H$ and $b \not\in H$, then $ab \not\in H$

*If $a \not\in H$ and $b \in H$, then $ab \not\in H$


Hence it is natural to ask when $a \not\in H$ and $b \not\in H$ implies $ab \in H$, ie. when a subgroup is a "parity subgroup", as in $G = \mathbb{Z}$ and $H = 2\mathbb{Z}$ or in $G = S_n$ and $H = A_n$. Suppose that $H$ is a proper subgroup since the case $H = G$ is not interesting. Then the following statements are equivalent:


*

*$[G:H] = 2$

*For all elements $a \not\in H$ and $b \not\in H$ of $G$, we have $ab \in H$.

*There exists a homomorphism $\phi: G \rightarrow \{1, -1\}$ with $\operatorname{Ker}(\phi) = H$.


So these parity subgroups are precisely all the subgroups of index $2$. I think you could use (2) in PatrickR's answer and (3) in DonAntonio's answer. Of course, they are both good and complete answers in their own, this is just one way I like to think about subgroups of index $2$.
