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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?

Thanks.

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  • $\begingroup$ 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
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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.

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