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Let $G=\langle x,y\mid x^2=y^2\rangle$, prove that exists subgroup of $G$ with index $2$.

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@Mariano: It says so. – Brian M. Scott Dec 12 '12 at 7:40
yes, this is my homework and i have no idea to prove that – Taukhtn Dec 12 '12 at 7:43
What does $G=<x,y|x^2=y^2>$ mean ? – Amr Dec 12 '12 at 12:59
Perhaps $G=<x,y|x^2=y^2>$ means the group generated by $x,y$ with the single relation $x^2=y^2$. – coffeemath Dec 12 '12 at 14:32
This group is also solvable, interesting, and also torsion free – Bhaskar Vashishth May 28 '15 at 17:59

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.

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Elegant. Or at least my kind of argument. :-) – Brian M. Scott Dec 13 '12 at 1:02

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 $G$.

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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.

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This is a good way to see things --- the abelianization. Especially when the latter can be seen easily, as in this example. +1 – coffeemath Dec 12 '12 at 17:11
I like your way - it is general, and a nice way to look at things...but it is kinda overkill! – user1729 Dec 12 '12 at 17:14

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