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 .


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Namaste, Claude Leibovici, Aweygan, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $f: \mathbb{Z}^2\rightarrow \mathbb{Z},\ f(x,y)=ay-bx$ $\endgroup$ – HK Lee May 21 '18 at 12:28
  • $\begingroup$ Great Sir .Thanks a Lot.... $\endgroup$ – MathLover May 21 '18 at 12:36

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 $2\mathbb Z v_1$.

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.