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$${ Z }_{ 3 }{ \times Z }_{ 4 }\simeq { Z }_{ 12 }$$ Above notations are ideal.

I tried $f(a,b)=4a+3b$ But, I run into a brick wall because of multiplication for homomorphism. How to prove it?

David Martínez
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YSRiemann
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3 Answers3

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NB: I could not entirely understand the question as posted, and here provide a proof that the two sides are isomorphic as additive groups.

I suggest that you write elements of $Z_3 \times Z_4$ as pairs $(a, b)$, where $a$ is in $Z_3$ and $b \in Z_4$. Then the simplest (group) isomorphism $f$ from $Z_{12}$ is $$ 1 \mapsto (1, 1) $$ You have to then ask whether $f(c) = 0$ for any value of $c$ other than $0$) (i.e., check that it's injective). Once you've done that, you know it's surjective, because the two sets have the same number of elements.

Let's check injectivity. For an element $c$, we have $$ f(c) = (c \bmod 3, c \bmod 4) $$ If that's $(0,0)$, then $c$ is a multiple of both $3$ and $4$; hence must be a multiple of 12 as well, and hence must be 0 in $Z_{12}$.

N.B.2:as @A.P. points out, this is also a homomorphism of rings, for \begin{align} f(cd) &= (cd \bmod 3, cd \bmod 4) \\ &= ((c \bmod 3) (d \bmod 3), (c \bmod 4) (d \bmod 4))\\ &= (c \bmod 3, c \bmod 4) \cdot (d \bmod 3, d \bmod 4) . \end{align}

John Hughes
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  • (a,a)(b,b)=(ab,ab)? – YSRiemann Jun 06 '15 at 10:41
  • Sorry...what I've proved is that they're isomorphic as additive groups. I wasn't claiming anything about multiplication. Was I supposed to? I couldn't quite make sense of the question. Are you hoping to show that they're isomorphic as rings? – John Hughes Jun 06 '15 at 10:43
  • @YSRiemann What is your question? This map is perfectly fine as an isomorphism of rings, too... – A.P. Jun 06 '15 at 10:47
  • AP: Good point. I should have said that, but since I was answering the group question, I thought I should be honest. :) – John Hughes Jun 06 '15 at 10:48
  • If you wish to show it explicitly, just observe that $([ab]_3,[ab]_4) = ([a]_3[b]_3, [a]_4[b]_4) = ([a]_3,[a]_4)([b]_3,[b]_4)$, where $[x]_n$ is the class of $x$ modulo $n$. – A.P. Jun 06 '15 at 10:50
  • A.P. AHH... Thank you! – YSRiemann Jun 06 '15 at 10:51
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How about you use $f : Z_{12} \to Z_4 \times Z_3$ such that $a \to (a\ \text{mod}\ 4, a\ \text{mod}\ 3)$

Bhaskar Vashishth
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Hint:

Use Bézouts identity $4-3=1$. The map is: $$(a\bmod 3,b\bmod 4)\mapsto 4a-3b\bmod 12.$$

Bernard
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  • Then, what is (a,b)(c,d)? Is it (ac,bd)? – YSRiemann Jun 06 '15 at 10:44
  • @YSRiemann This map is the inverse of John Hughes's map. To show that it is a ring morphism you need to show that $(4a - 3b)(4c - 3d) \equiv 4ac - 3bd \pmod{12}$. – A.P. Jun 06 '15 at 10:56