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Is $(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$,

where $4\mathbb Z\times 5 \mathbb Z$ is given as a subgroup of $\mathbb Z\times \mathbb Z$?

My feelings says that I need to find well defined $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which has a kernel $4\mathbb Z\times 5\mathbb Z$

So using first isomorphism theorem I can show the isomorphism.

I could not find the map and how can I show it in different way?

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  • $\begingroup$ Are you regarding these mathematical concepts as groups or as rings? (I might be able to help with the former; not so much, the latter.) $\endgroup$
    – Shaun
    Jan 18, 2021 at 17:00
  • $\begingroup$ yeah this about group thats why I wanted to prove it with 1st isom. theorm. $\endgroup$ Jan 18, 2021 at 17:00
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    $\begingroup$ Rings have their isomorphism theorems too $\endgroup$
    – Shaun
    Jan 18, 2021 at 17:02
  • $\begingroup$ Editted: Where $4\mathbb Z\times 5 \mathbb Z$ is given as a subgroup of $\mathbb Z\times \mathbb Z$ $\endgroup$ Jan 18, 2021 at 17:04
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    $\begingroup$ The fact that the subgroup is generated by two elements should, according to the third isomorphism theorem, make you suspect that the rank (i.e. the minimal number of order-infinite generators) of $2$ ought to decrease by two as you take the quotient. $\endgroup$
    – Arthur
    Jan 18, 2021 at 17:28

6 Answers 6

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By definition of the quotients:

\begin{alignat}{1} (\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z) &= \{(m,n)+4\Bbb Z\times5\Bbb Z\mid m,n\in\Bbb Z\} \\ &= \{(m+4\Bbb Z,n+5\Bbb Z)\mid m,n\in\Bbb Z\} \\ &= \{(k,l)\mid k\in\Bbb Z/4\Bbb Z,l\in\Bbb Z/5\Bbb Z\} \\ &= \Bbb Z/4\Bbb Z \times\Bbb Z/5\Bbb Z \\ \end{alignat}

So, your quotient is precisely (not just isomorphic to) $\Bbb Z/4 \Bbb Z\times\Bbb Z/5\Bbb Z $, which is not isomorphic to $\Bbb Z$, if only for reasons of cardinality.

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No.

Since $$G:=\Bbb Z\times \Bbb Z\cong\langle a,b\mid ab=ba\rangle,$$

we can write the subgroup $H:=4\Bbb Z\times 5\Bbb Z$ as the normal subgroup $$\langle \langle a^4,b^5\rangle \rangle, $$ so that

$$\begin{align} G/H &\cong \langle a,b\mid a^4, b^5, ab=ba\rangle\\ &\cong \langle a,b,c\mid a^4, b^5, ab=ba, a=cb^{-1}\rangle \\ &\cong \langle b,c\mid (cb^{-1})^4, b^5, c=b(cb^{-1})\rangle\\ &\cong\langle b,c\mid c^4=b^4, b^5, cb=bc\rangle\\ &\cong\langle b,c\mid c^4=b^{-1}, b^5, cb=bc\rangle \\ &\cong \langle b,c\mid b=c^{-4}, (c^{-4})^5, c^{-3}=c^{-3}\rangle \\ &\cong\langle c,d\mid d=c^{-1}, d^{20}\rangle \\ &\cong\langle d\mid d^{20}\rangle\\ &\cong\Bbb Z/ 20\Bbb Z\\ &\not\cong\Bbb Z. \end{align}$$

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  • $\begingroup$ Using group presentations is somewhat 'advanced' in my opinion. I'm not sure if it is helpful to the OP. $\endgroup$
    – J. De Ro
    Jan 18, 2021 at 18:12
  • $\begingroup$ I was thinking that, @QuantumSpace, when I was halfway through the calculation, but then I figured it wouldn't hurt to have a different perspective. $\endgroup$
    – Shaun
    Jan 18, 2021 at 19:23
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    $\begingroup$ Yes, it may benefit more advanced readers for sure :) $\endgroup$
    – J. De Ro
    Jan 18, 2021 at 19:31
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    $\begingroup$ Fond of the way you took the presentation of the group. 👌 $\endgroup$
    – Mikasa
    Jan 19, 2021 at 7:15
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What is the kernel of the group homomorphism $\mathbb Z \times \mathbb Z \rightarrow \mathbb Z/4\mathbb Z \times \mathbb Z/5\mathbb Z$ given by

$$(x,y) \mapsto (x+4\mathbb Z, y +5\mathbb Z)?$$

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Consider the map $$\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}: (x,y) \mapsto (x+ 4 \mathbb{Z}, y + 5\mathbb{Z})$$ This map is surjective and has kernel $\mathbb{4}\mathbb{Z}\times 5\mathbb{Z}$, so by the first isomorphism theorem, $$(\mathbb{Z}\times \mathbb{Z})/(4\mathbb{Z}\times 5 \mathbb{Z})\cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\cong \mathbb{Z}/20\mathbb{Z}$$

The latter group contains $20$ elements and is finite, so the group you are interested in is finite. Hence, it cannot be isomorphic to $\mathbb{Z}$.

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There shouldn't be such a map. You may want to check for elements with finite order in both groups.

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    $\begingroup$ OP is treating the objects as groups. $\endgroup$ Jan 18, 2021 at 17:04
  • $\begingroup$ @DonThousand Do you agree now? $\endgroup$
    – TomTom314
    Jan 18, 2021 at 17:10
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There are a number of answers explaining why your isomorphism isn't true, but I am going to try to answer to the question maybe you wanted yo ask: let $A = \mathbb{Z}\times \mathbb{Z}$, and $B = \{(4x,5x)\in A\mid x\in \mathbb{Z}\}$. Note that $B$ is not the subgroup $4\mathbb{Z}\times 5\mathbb{Z}$, and is not even isomorphic to it since it is instead isomorphic to $\mathbb{Z}$.

Then indeed $A/B\simeq \mathbb{Z}$, because $4$ and $5$ are coprime. Explicitly, the surjective morphism $A\to \mathbb{Z}$ given by $(a,b)\mapsto 5a-4b$ has kernel $B$.

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