isomorphic

**apple2009** · November 21st 2009, 08:53 PM

Let $G=\mathbb{Z}_4\times\mathbb{Z}_3$ . Show that $G$ is isomorphic to $\mathbb{Z}_{12}$ .

**Jose27** · November 21st 2009, 09:45 PM

Take $(1,1)\in \mathbb{Z}_4 \times \mathbb{Z}_3$ then let $k\in \mathbb{N}$ and we have $(k1,k1)=0$ iff $4\vert k$ and $3\vert k$ the least such number is $[4,3]$ (the min. comm. mult.) and it's well known that $[a,b](a,b)=ab$ then since $(4,3)=1$ ...

**qmech** · November 22nd 2009, 11:05 AM

The map
Z12 -> G
defined by n -> (n mod 4, n mod 3 ) is 1:1 and onto.

Alternatively,
G-> Z12
defined by
(a,b) -> 3*a + 4*b mod 12
should also be 1:1 and onto.

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