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In previous questions, I've proved that $B$ is a free abelian group of rank $2$. Then naturally $B$ is isomorphic to $\Bbb Z^2$, right? Then $\mathbb{Z}^2/B$ is isomorphic to $\mathbb{Z}^2/\mathbb{Z}^2$.

Isn't this just the trivial group?

How am I suppose to express it as a direct product?

Shaun
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everwith
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    If $A=\mathbb{Z}$, and $B=2\mathbb{Z}$, then $B$ is "naturally" isomorphic to $\mathbb{Z}$. Is $A/B=\mathbb{Z}/2\mathbb{Z}$ isomorphic to $\mathbb{Z}/\mathbb{Z}$? (Not sure what you mean by "naturally isomorphic" as usually 'natural' has a technical meaning, and it is unlikely that this is what you have here; hence the scare quotes) – Arturo Magidin Aug 10 '22 at 21:44
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    A free abelian group of rank two is isomorphic to $\mathbb{Z}^2$, however $\mathbb{Z}^2$ has proper subgroups isomorphic to itself. – CyclotomicField Aug 10 '22 at 21:49
  • @CyclotomicField Doesn't the Fundamental Theorem of Abelian Groups guarantees that $\mathbb{Z}^2$ cannot be partitioned? Also, even if $\mathbb{Z}^2$ does have proper isomorphic subgroups, it doesn't change the fact that $\mathbb{Z}^2/\mathbb{Z}^2$ gives the trivial group? – everwith Aug 10 '22 at 21:54
  • @everwith you're right about $\mathbb{Z}^2 / \mathbb{Z}^2$ but that's just one case. You have have some other quotients to consider. – CyclotomicField Aug 10 '22 at 22:04

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Observe that $\Bbb Z\cong 2\Bbb Z\cong 3\Bbb Z$.

Let $G=\langle a,b\mid ab=ba\rangle\cong \Bbb Z^2$.

Suppose

$$\begin{align} B&:=\langle a^2,b^2\rangle_G \\ &\cong 2\Bbb Z\times 2\Bbb Z\\ &\cong \Bbb Z^2. \end{align}$$

Then

$$\begin{align} G/B&\cong \langle a,b\mid ab=ba\rangle/\langle a^2,b^2\rangle_G\\ &=\langle a,b\mid a^2, b^2, ab=ba\rangle\\ &\cong \Bbb Z_2\times \Bbb Z_2. \end{align}$$

But if

$$\begin{align} H&:=\langle a^2,b^3\rangle_G \\ &\cong 2\Bbb Z\times 3\Bbb Z\\ &\cong \Bbb Z^2, \end{align}$$

then

$$\begin{align} G/H&\cong \langle a,b\mid ab=ba\rangle/\langle a^2,b^3\rangle_G\\ &=\langle a,b\mid a^2, b^3, ab=ba\rangle\\ &\cong \Bbb Z_2\times \Bbb Z_3. \end{align}$$

Shaun
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  • Just want to confirm, $\langle a,b\mid a^2, b^2, ab=ba\rangle$ = ${ x\langle a^2\rangle\mid x\in \langle a\rangle}\times { y\langle b^2\rangle\mid y\in \langle b\rangle}$, right? – everwith Aug 10 '22 at 23:22
  • No, @everwith; here $\langle x, y\rangle_G$ is the subgroup of $G$ generated by $x$ and $y$. Basically, it is the subgroup given by all products in $G$ of $x$ and $y$ only. – Shaun Aug 11 '22 at 10:55
  • Also $$\langle a, b\mid a^2, b^2, ab=ba\rangle\cong \langle a\mid a^2\rangle \times\langle b\mid b^2\rangle.$$ – Shaun Aug 11 '22 at 11:44
  • See here, @everwith. – Shaun Aug 11 '22 at 11:45