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Suppose $G = \langle x,y\mid x^3 = y^5\rangle$. How can I compute the abelianization of such a group?


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Are you familiar with Smith normal form? – Qiaochu Yuan Nov 27 '12 at 18:30
No I am not, but I will check thank you! – Ferenc Nov 27 '12 at 18:33
Consider adding a new element $z=x^3=y^5$. By commuting terms around you should be able to present every element of the abelianization in the form $x^py^q$, and then in the form $x^iy^jz^k$ with $0\leq i\lt 3$ and $0\leq j\lt 5$. – Steven Stadnicki Nov 27 '12 at 18:44
up vote 0 down vote accepted

When you add the relation $xy = yx$ then all elements can be written in the form $x^ay^b$ with $a, b \in \mathbb{Z}$. Moreover $x^ay^b = e$ precisely when $(a, b)$ is a multiple of $(3, -5)$. This means that your group is isomorphic to $\mathbb{Z}^2 / \mathbb{Z}(3, -5) \cong \mathbb{Z}$. (The "precisely" might need some explaining.)

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Thanks! Why is $\mathbb Z^2/\mathbb Z(3,-5)\cong \mathbb Z$? Should not it be isomorphic to $\mathbb Z/3 \oplus \mathbb Z/5$? – Ferenc Nov 27 '12 at 18:52
@Ferenc: This is because $\gcd(3,5) = 1$. In this case you have for example $\mathbb{Z}^2 = \mathbb{Z}(3,-5) \oplus \mathbb{Z}(1, -2)$. – WimC Nov 27 '12 at 18:55
I see! Thank you! – Ferenc Nov 27 '12 at 18:59

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