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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?

  • That's the best way, certainly. – Ian Coley Oct 23 '13 at 06:48
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    You already found the generator, which should be in both. Right? You pretty much have the problem solved. – MaybeALlama Oct 23 '13 at 06:53
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    In fact Z mod m X Z mod n is isomorphic to Z mod mn when (m,n) =1. This can be proved. – wannadeleteacct Oct 23 '13 at 07:01
  • How do I show it's a bijective homomorphism? –  Oct 23 '13 at 07:08
  • Is not an isomorphism defined as bijective homomorphism? – wannadeleteacct Oct 23 '13 at 07:16
  • @Manasi : Not in an intention to point out some error but, Isomorphism is not "defined" as bijective homomorphism (i mean not in all categories).. for example in topological spaces Isomorphism(homeomorphism) of two spaces is not just bijective continuous function.... (excuse me if you are not familiar with topological spaces).. I was in same feeling sometime back and somebody from this forum pointed ut my wrong idea and i thought that could help you for better understanding.. –  Oct 23 '13 at 07:30
  • Thanks for the added information, however I am pretty much sure that in the category of groups,an isomorphism is a bijective homomorphism. An isomorphism is an invertible morphism in the category of groups. I was just thinking about groups when I posted and not in a general context.( Yes, I am weak at Topology! Hence, I have not thought about Categories and all that in a long time :( ) – wannadeleteacct Oct 23 '13 at 07:47

2 Answers2

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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 $\mathbb{Z}/12\cong \mathbb{Z}\oplus \mathbb{Z}/(3\mathbb{Z}\oplus 4\mathbb{Z})$.

Moss
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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.

Prahlad Vaidyanathan
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  • This isn't an isomorphism is it? 2,1 and 1,2 go to 2. –  Oct 23 '13 at 07:17
  • 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. – wannadeleteacct Oct 23 '13 at 07:22
  • 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. –  Oct 23 '13 at 07:34
  • 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? – wannadeleteacct Oct 23 '13 at 07:52
  • Unless I'm going to write out all 12 mappings, I still don't follow the pattern here. –  Oct 23 '13 at 08:10
  • The idea is to get a map f(x,y) = ax + by. I wrote all the 12 cases and finally figured it out by considering f(1,0) and f(0,1). The map is f(x,y) = 4x + 9y. – wannadeleteacct Oct 23 '13 at 09:40