2

After reading this and this and I’m wondering if there’s a shorter and more intuitive way: would it be correct to claim the following?

By change of basis to $(1,0), (1,1)$ we can write \begin{align*} (\mathbb{Z} \oplus \mathbb{Z}) / \langle (2, 2) \rangle &\cong \Bigl( \langle (1,0) \rangle \oplus \langle (1,1) \rangle \Bigr) \Big/ \Bigl( \langle (0,0) \rangle \oplus \langle (2,2) \rangle \Bigr) \\ &\cong \Bigl( \langle (1,0) \rangle \big/ \langle (0,0) \rangle \Bigr) \oplus \Bigl( \langle (1,1) \rangle \big/ \langle (2,2) \rangle \Bigr) \\ &\cong \mathbb{Z} \oplus \mathbb{Z}_2 \end{align*} where the first isomorphism is by change of basis for $\mathbb{Z}^2$ and the second is by the fact that the quotient of direct sums of abelian groups is the direct sums of the quotients since $\langle (0,0) \rangle$ and $\langle (2,2) \rangle$ are subgroups of the respective groups in the product (as seen here).

If not – is there a similar way to state this? I find these transitions more intuitive than the Smith normal form or finding appropriate isomorphisms.

Jendrik Stelzner
  • 17,081
Anon
  • 1,965

1 Answers1

5

This depends on how comfortable you are with group presentations.

One way to write $G=\Bbb Z\times \Bbb Z$ is as the group given by the presentation

$$G\cong\langle a,b\mid ab=ba\rangle$$

and the subgroup $H=\langle (2,2)\rangle$ can then be written as the normal subgroup $\langle (ab)^2\rangle$ generated by $(ab)^2$ (because you can think of $a$ as $(1,0)$ and $b$ as $(0,1)$). Now

$$\begin{align} G/H&\cong \langle a,b\mid ab=ba\rangle/\langle (ab)^2\rangle\\ &\cong\langle a,b\mid (ab)^2, ab=ba\rangle\\ &\cong\langle a,b,c\mid (ab)^2, ab=ba, c=ab\rangle\\ &\cong\langle a,b,c\mid c^2, c=ba, b=a^{-1}c\rangle\\ &\cong\langle a,c\mid c^2, c=(a^{-1}c)a\rangle\\ &\cong\langle a,c\mid c^2, ac=ca\rangle\\ &\cong \Bbb Z\times \Bbb Z_2 \end{align}$$

by Tietze transformations.

Shaun
  • 47,747