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I wanted to find group isomorphic to $(\Bbb Z \oplus \Bbb Z )/\langle(4,2)\rangle $.
I don't know how to calculate that .I think it may be isomorphic to $(\Bbb Z_4 \oplus \Bbb Z )$ or $(\Bbb Z_2 \oplus \Bbb Z )$ From Hint provided Below :
$(\Bbb Z \oplus \Bbb Z )$= $(\Bbb Z(2,1) \oplus \Bbb Z(1,0) )$ and $\langle (4,2) \rangle $=$2\langle (2,1)\rangle$
So $(\Bbb Z \oplus \Bbb Z )/\langle(4,2)\rangle $=$(\Bbb Z(2,1) \oplus \Bbb Z(1,0) )$/$2\langle (2,1)\rangle$= $(\Bbb Z_2 \oplus \Bbb Z )$
Is this right?

Any Help will be appreciated .

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closed as off-topic by Derek Holt, Namaste, Claude Leibovici, Aweygan, José Carlos Santos May 21 '18 at 23:13

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Define $$f:\mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}_2,\ f(x,y)=(x-2y,y)$$

Here $$ f((x,y)+(a,b)) =(x+a-2y-2b,y+b) =f((x,y))+f((a,b)) $$ is a group homomorphism.

And the kernel is $( 4t,2t)$ so that $\mathbb{Z}\oplus \mathbb{Z}_2$ is desired.

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  • $\begingroup$ Sir Very Elegant Solution .I now understand to use factor group to define homomorphism .But One doubt is that I can go in following manner I know kernel and domain so I can find homomorphism .But I am not able to convince my self about Range Why $Z \times Z_2$? Please Help me out $\endgroup$ – MathLover May 21 '18 at 11:56
  • $\begingroup$ Note that by $x$, $(x-2,1),\ (x-0,0)$ represent any element in $\mathbb{Z}\times \mathbb{Z}_2$. $\endgroup$ – HK Lee May 21 '18 at 12:00
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    $\begingroup$ Thanks a lot Sir .That means for kernel (kb,k) homomorphism will be f(x,y)=(x-ky,y) and Isomorphic to $Z \times Z_k$ .Is this possible to only (kb,k) form or also true in case of any (a,b) where a and b may be any thing even relatively prime ? $\endgroup$ – MathLover May 21 '18 at 12:05
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    $\begingroup$ $f: \mathbb{Z}^2\rightarrow \mathbb{Z},\ f(x,y)=ay-bx$ $\endgroup$ – HK Lee May 21 '18 at 12:28
  • $\begingroup$ Great Sir .Thanks a Lot.... $\endgroup$ – MathLover May 21 '18 at 12:36
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Hint: Take $v_1=(2,1)$ and $v_2=(1,0)$. Then $\mathbb Z^2 = \mathbb Z v_1 \oplus \mathbb Z v_2$ and your subgroup is $2\mathbb Z v_1$.

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  • $\begingroup$ Sir ,Great Solution .But for me choice of $v_1 , v_2$ is very unnatural .In sense that if I have any other problem I can not choose this directly .Is this with practice I get or is there any algorithm for choice of this $v_1 , v_2$.Thanks A lot for Reply $\endgroup$ – MathLover May 21 '18 at 11:44
  • $\begingroup$ @SRJ, ask a separate question for the general case $(\Bbb Z \oplus \Bbb Z )/\langle(a,b)\rangle$. $\endgroup$ – lhf May 21 '18 at 11:46
  • $\begingroup$ @SRJ, the general algorithm is the Smith normal form. $\endgroup$ – lhf May 21 '18 at 11:47

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