Prove that group $G$ has subgroup with index 2 Let $G=\langle x,y\mid x^2=y^2\rangle$, prove that exists subgroup of $G$ with index $2$.
 A: The relation $x^2=y^2$ is the same as $xxy^{-1}y^{-1}=1.$ If a group $G$ has a presentation in which all relations have even length, then there is a well defined length mod 2 of any $w \in G$, since the process of reducing a word by cancelling $tt^{-1}$ does not change the length of $w$ mod 2, and use of the relations does not change the length mod 2 either, since the relations have even length. So there is a homomorphism $f:G \to Z_2$ taking $w \in G$ to its length mod 2. The kernel $K$ of $f$ is then a subgroup of index 2.
A: Observe that $x^2=y^2$ is a central element of $G$ (i.e. it belongs to the center of $G$). If we quotient out we have $G/\langle x^2 \rangle = \langle a,b\ |\ a^2=b^2=1 \rangle$ (here $a = x \langle x^2 \rangle$ and $b= y \langle x^2 \rangle$). Now this is the infinite Dihedral group, and its subgroup $\langle ab \rangle$ has index $2$. Indeed $b^{-1} (ab) b = ba = (ab)^{-1} \neq ab$, so $\langle ab \rangle \cap \langle b \rangle = \{1\}$, $\langle ab \rangle \unlhd G$ and $G/ \langle ab \rangle = \langle ab \rangle \langle b \rangle / \langle ab \rangle \cong \langle b \rangle / \langle ab \rangle \cap \langle b \rangle \cong \langle b \rangle \cong C_2$.
Therefore $\langle xy,x^2 \rangle$ has index $2$ in $G$.
A: Look at the abelianisation of your group. It is the Klein $4$-group. This maps onto the cyclic group of order two. Thus, your group maps onto the cyclic group of order two. The kernel of this map has index two.
In general, a group $G$ has a subgroup of index two if and only if $G$ maps onto the cyclic group of order two (because subgroups of order two are always normal), and the easiest way to check this is to look at the abelianisation.
