I have the dihedral group $$D_n =\langle x,y\mid x^n, y^2,(xy)^2\rangle$$ I need to show that the $D_n$ abelianised is isomorphic to $Z_2$ if $n$ is odd and $Z_2 \oplus Z_2$ if $n$ is even. How do I show this?
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Note that the abelianization is generated by $y$ and $xy$, both of which have order $2$. – Andrew Dudzik May 07 '15 at 19:07
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The abelianisation is $\langle x,y\mid x^n, y^2, (xy)^2,xy=yx\rangle$. From the relations, you get $1 = (xy)^2 = x^2y^2 = x^2$. If $n$ is even, then the relator $x^n$ is redundant while, for odd $n$, you can conclude that $x=1$ and eliminate that generator. Thus, we end up with either $\langle y\mid y^2\rangle$ or $\langle x,y\mid x^2, y^2, xy=yx\rangle$.
James
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I do not understand why when n is even that the relator $x^n$ is redundant? – Al jabra May 07 '15 at 19:35
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Suppose that $n =2k$. You have $x^2 = 1$ and $x^{2k}=1$. But $x^{2k}=1$ follows from $x^2=1$ because $x^{2k} = (x^2)^k$. – James May 07 '15 at 22:11