Let $G$ be a group generated by two involutions $x$ and $y$. Show that there is some $N\unlhd G$ with $[G:N]=2$. This is Exercise 3.35 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search, it is new to MSE.
The Details:
Presentations are not covered in the book so far, so, presumably, there is a way to answer it without thinking of $G$ given by (some quotient of) the free product
$$\Bbb Z_2\ast\Bbb Z_2\cong \langle x,y\mid x^2,y^2\rangle.$$
The combinatorial-group-theory tag does not apply.
Since I sometimes get involutions and idempotents mixed up, here is a

Definition: An element $f$ of a group $G$ with identity $e$ is an involution if $f^2=e$.

Since there is a plethora of definitions for a normal subgroup, the following is from the book cited above.

Definition 2: A subgroup $H$ of a group $G$ is normal in $G$, written $H\unlhd G$, if $$aH=Ha$$ for all $a\in G$.

The Question:

Let $G$ be a group generated by two involutions $x$ and $y$. Show that $G$ has a normal subgroup of index two.

Thoughts:
It won't help to consider, without loss of generality, that the normal subgroup - let's call it $N$ - is $\langle x\rangle$, say, since, although $x\langle x\rangle=\langle x\rangle x$, trivially, the same cannot be said in general for $y\langle x\rangle$ and $\langle x\rangle y$.
It does give me the idea, though, that, to test the normality of $N$ in $G$, we need only show that $xN=Nx$ and $yN=Ny$.
Exactly where the $[G:N]=2$ comes from is a mystery to me. My hope is that some pithy choice of generators for $N$ would work, but, well, that's using techniques not yet covered in the book.
I feel like this is a question I ought to be able to answer myself. I've given it a few days and this is all I have.
Please help :)
 A: Consider $\langle xy\rangle$. Note that $(xy)^{-1}=yx$. Therefore, $x(xy)x^{-1} = xxyx = yx = (xy)^{-1}\in\langle xy\rangle$, and $y(xy)y^{-1} = y(xy)y = yx = (xy)^{-1}\in\langle xy\rangle$. Thus, $\langle xy\rangle$ is normal. Call it $N$.
Now note that $G/N$ is abelian, since $[x,y]=xyxy\in N$ and $G/N$ is generated by $xN$ and $yN$. On the other hand, if $x\in N$ then there exists $n\in\mathbb{Z}$ such that $x=(xy)^n$, and since $x$ has order $2$ we may assume $n\gt 0$. Pick $n$ minimal with this property. Then $x=x(yx)^{n-1}y$, so $(yx)^{n-1}y=e$. Therefore $(yx)^{n-1}=y$, so $y(xy)^{n-2}x = y$. Therefore, $(xy)^{n-2} = x$. By the minimality of $n$, either $n=1$ or $n=2$. If $n=1$, then $y=(yx)^0=e$, which contradicts the assumption that $y$ has order $2$. Therefore, $n=2$. But then $yx
=y$, so $x=e$, again a contradiction. Thus $x\notin N$. Symmetrically, $y\notin N$. Thus, $G/N$ is abelian, nontrivial, generated by two elements of order $2$. But since $xN=yN$ (as $y=xxy\in xN$), $G/N$ is cyclic of order $2$. So $N$ will work.
A: Let $X=\mathbb{R}P^2\vee \mathbb{R}P^2$ be the wedge sum of two copies of $\mathbb{R}P^2$. Consider the continuous function $f:X\to \mathbb{R}P^2$ which is identity on one copy and is constant on the other copy. Then $$f_{*}:G=\pi_1(X)\to \pi_1(\mathbb{R}P^2)=\mathbb{Z}/2\mathbb{Z} $$ is a surjective homomorphism whose kernel is a subgroup of index $2$.
Remark: This homomorphism is actually the homomorphism $\phi$ which is presented in the comment to OP post.
