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I am reading these lecture notes and they suggest the following generalisation of a specific example for $\mathbb Z_2 \times \mathbb Z_3 \cong Z_6 $:

There exists an isomorphism between $\mathbb Z_n \times \mathbb Z_m$ and $\mathbb Z_{nm}$ if $\gcd(n,m)=1$.

Clearly these two groups are of the same cardinality, since we can choose $n$ elements for the first coordinate and $m$ for the second. This is precisely the number of elements in $\mathbb Z _{nm}$.

Secondly, I know that $1$ generates the second group and that $(1,1)$ generates the first group. Is it as simple to just identify powers with each other, is there a justification to do so besides the fact that both elements have order $mn$ in their own group? So I would suggest: $$f((1,1)) = 1$$ as a starting point. And then $$f((a,a)) = a \cdot 1$$ where $1 \leq a \leq n \cdot m$

I suppose that since$$ \gcd(n, m) \cdot lcm(n,m)= nm $$and since $\gcd(n,m)=1$ , we know that $lcm(n,m)=nm$. This would indeed mean that the order of the element (1, 1) is $nm$. I feel like I a missing a step in the reasoning here. I suppose that since the least common multiple is $nm$ there are no "overlapping" powers. Can anybody help me make this more concrete?

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    You will find useful the well known chinese remainder theorem for ideals. Here a link https://math.stackexchange.com/questions/1102037/the-chinese-remainder-theorem-for-rings – DrinkingDonuts Feb 18 '19 at 21:29
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    How do you know $(1,1)$ generates the first group? This is false for $\mathbb Z_2\times\mathbb Z_2$. – Cheerful Parsnip Feb 18 '19 at 21:52
  • Precisely because $m \neq n$ I think this should be a condition, that's probably why they have used different letters. The notes I am using forgot to mention that. –  Feb 18 '19 at 21:56
  • $(1,1)$ worked for $\mathbb Z_2 \times \mathbb Z_3$ and as a hint these notes suggested to use this example to generalise it for other numbers that are coprime. –  Feb 18 '19 at 21:57
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    The hypothesis that $n\ne m$ is redundant. Since $\gcd(n,m)=1$, you have $n=m$ only when both are $1$ and the theorem is true in that case. – egreg Feb 18 '19 at 23:23

1 Answers1

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Working with isomorphisms is simpler if you use homomorphisms.

There is a homomorphism $\varphi\colon\mathbb{Z}\to\mathbb{Z}_n\times\mathbb{Z}_m$ defined by $\varphi(k)=([k]_n,[k]_m)$ (where $[k]_a$ denotes the residue class of $k$ modulo $a$).

What's the kernel? Clearly $k\in\ker\varphi$ if and only if both $n$ and $m$ divide $k$, so it is the subgroup generated by the lowest common multiple between $n$ and $m$. This is $nm$ when $\gcd(n,m)=1$. Thus we have an injective homomorphism $$ \mathbb{Z}/\!\ker\varphi=\mathbb{Z}_{nm}\to\mathbb{Z}_n\times\mathbb{Z}_m $$ defined by $[k]_{mn}\mapsto([k]_n,[k]_m)$, using the first homomorphism theorem.

Now count elements.

egreg
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  • They are of the same cardinality, thus this map is bijective thus it is an isomoprhism! cool, thanks! –  Feb 19 '19 at 09:49
  • Reasoning the other way makes a whole lot of sense ^^ –  Feb 19 '19 at 12:20
  • How come the map is well-defined though? –  Feb 23 '19 at 22:39
  • @WesleyStrik The homomorphism theorem applies. – egreg Feb 23 '19 at 22:57
  • I mean, why is $\phi (k) = \phi(l)$ whenever $k \equiv l \mod{mn}$ –  Feb 23 '19 at 23:03
  • @WesleyStrik Because the homomorphism theorem tells you it. – egreg Feb 23 '19 at 23:05
  • A homomorphism is a map with the property $f(ab)=f(a)f(b)$, I see how this property holds, but I do not see why it is a map. In order to use the homomorphism theorem, would you not first have to show it is a homomorphism? https://en.wikipedia.org/wiki/Fundamental_theorem_on_homomorphisms –  Feb 23 '19 at 23:29
  • @WesleyStrik $\varphi$ is a homomorphism, isn't it? Its kernel is $mn\mathbb{Z}$. If $f\colon G\to G'$ is a homomorphism of additive groups and $x-y\in\ker f$, then $f(x)=f(y)$. So $f$ induces a injective homomorphism $\bar{f}\colon G/!\ker f\to G'$. This is the homomorphism theorem. – egreg Feb 23 '19 at 23:31
  • I was thinking more elementary, you're a couple of steps ahead. It actually follows from a simple divisibility argument that $\phi$ is well defined and thus a map, and then one can make the step to homomorphism :) –  Feb 23 '19 at 23:34