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How can I show that $\mathbb{Z}_3\times\mathbb{Z}_4$ is isomorphic to $\mathbb{Z}_{12}$ I found an order twelve generator, and that was the hint, but can I show it is an isomorphism without showing it's a bijective homomorphism, or is that the only way?

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  • $\begingroup$ That's the best way, certainly. $\endgroup$ – Ian Coley Oct 23 '13 at 6:48
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    $\begingroup$ You already found the generator, which should be in both. Right? You pretty much have the problem solved. $\endgroup$ – Steven Walton Oct 23 '13 at 6:53
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    $\begingroup$ In fact Z mod m X Z mod n is isomorphic to Z mod mn when (m,n) =1. This can be proved. $\endgroup$ – wannadeleteacct Oct 23 '13 at 7:01
  • $\begingroup$ How do I show it's a bijective homomorphism? $\endgroup$ – user82004 Oct 23 '13 at 7:08
  • $\begingroup$ Is not an isomorphism defined as bijective homomorphism? $\endgroup$ – wannadeleteacct Oct 23 '13 at 7:16
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Alternatively, you could try to calculate the kernel of the homomorphism $\mathbb{Z}\oplus\mathbb{Z}\rightarrow \mathbb{Z}/12$ given by $(x,y)\mapsto 4x+3y$. With any luck, the kernel will turn out to be $3\mathbb{Z}\oplus 4\mathbb{Z}$.

Then apply the first isomorphism theorem to say that $\mathbb{Z}/12\cong \mathbb{Z}\oplus \mathbb{Z}/(3\mathbb{Z}\oplus 4\mathbb{Z})$.

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Just define $f : \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/12\mathbb{Z}$ by $$ f([a],[b]) = [ab] $$ You need to check that this is well-defined (ie. if $a \equiv c\pmod{3}$ and $b\equiv d\pmod{4}$, then $ab\equiv cd\mod{12}$). This takes a little work, but once that is done, it is clear that it is the isomorphism you are looking for.

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  • $\begingroup$ This isn't an isomorphism is it? 2,1 and 1,2 go to 2. $\endgroup$ – user82004 Oct 23 '13 at 7:17
  • $\begingroup$ Map the identity of $\mathbb{Z}_3\times\mathbb{Z}_4$ to the identity of $\mathbb{Z}_{12} $. Then try your luck in defining the map. $\endgroup$ – wannadeleteacct Oct 23 '13 at 7:22
  • $\begingroup$ I am afraid I still don't follow. Mapping 0,0 to 0, would map things like 1,2 and 2,1 to 3, which is not injective. $\endgroup$ – user82004 Oct 23 '13 at 7:34
  • $\begingroup$ Not necessarily. That is not the map I had in mind.Now,map the generator (1,1) of $\mathbb{Z}_3\times\mathbb{Z}_4$ to the generator 1 of $\mathbb{Z}_{12}$. Does this help? $\endgroup$ – wannadeleteacct Oct 23 '13 at 7:52
  • $\begingroup$ Unless I'm going to write out all 12 mappings, I still don't follow the pattern here. $\endgroup$ – user82004 Oct 23 '13 at 8:10

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