I found that $(\mathbb Z_4\times\mathbb Z_6)/\langle(2,3)\rangle$ should be isomorphic to $\mathbb Z_4\times\mathbb Z_3$.

But I cannot construct a homomorphism $\phi:\mathbb Z_4\times\mathbb Z_6\to\mathbb Z_4\times\mathbb Z_3$ such that $\ker\phi=\langle(2,3)\rangle$.

A function such that $(a,b)\to(a,b)$ when $b\leq2$ and $(a,b)\to(a+2,b)$ otherwise seems like the function I want, but it's formula is too complicated.

Can anyone help me construct $\phi$ in efficient manner?


  • $\begingroup$ How did you "find" the isomorphism if you don't know the homomorphism explicitly, in the first place? $\endgroup$ – Kushal Bhuyan Jun 6 '16 at 11:59
  • $\begingroup$ @KushalBhuyan Actually I'm reading Fraleigh 15.11 Example, and in here : The quotient must be order 12, so it must be isomorphic to $\mathbb Z_4\times\mathbb Z_3$ or $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_3$. Since there is an element of order 4, the quotient should be isomorphic to $\mathbb Z_4\times\mathbb Z_3$. $\endgroup$ – Jinmu You Jun 6 '16 at 12:04
  • $\begingroup$ Oh can you give me the page no. I want to take a look too. For the moment try this $\phi (m,n)=(m+2, n(mod 3))$, I am not sure though. $\endgroup$ – Kushal Bhuyan Jun 6 '16 at 12:08
  • $\begingroup$ @KushalBhuyan Abstract Algebra of Fraleigh, 7ed, pg.147. I'll try that. $\endgroup$ – Jinmu You Jun 6 '16 at 12:10
  • $\begingroup$ @KushalBhuyan That function carries $(2,0)\to (0,0)$, which is not an element of the given kernel. $\endgroup$ – Jinmu You Jun 6 '16 at 12:12

Your formula looks right to me except the second number in the target needs to be in the range $0$ to $2$. One way to express it is $(a,b) \to (a+2(b/3), b \bmod 3)$ where the divide is integer divide.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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