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Let $G$ be an abelian group which is killed by multiplication with the integer $n\geq 1$.

Let $n=a\cdot b$ with $a,b \geq 1$ and relatively prime.

Denote by $G[a]$ resp. $G[b]$ the $a$-resp. $b$-torsion subgroups of $G$.

Then why are the following maps isomorphisms:

(1) $G[a] \times G[b] \rightarrow G$ (addition map)

(2) $G \rightarrow G[a]\times G[b], g\mapsto (b\cdot g,a\cdot g)$ ?

Note that I don't assume finiteness of $G$ and that injectivity of both maps are clear to me. The problem is surjectivity.

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How do you know that these maps are isomorphisms? –  Rasmus Sep 13 '12 at 19:11
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2 Answers

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Find $u,v\in\mathbb Z$ such that $u a + v b=1$.

(1) For $g\in G$, let $x=ua\cdot g$, $y=vb\cdot g$. Then $x+y=g$ and $bx=un\cdot g=0$, $ay=vn\cdot g$. Hence $(y,x)\mapsto g$.

(2) For $(x,y)\in G[a]\times G[b]$, let $g=vx+uy$. Then $b\cdot g = vbx+0=(vb+ua)x=x$ and similary $a\cdot g=y$.

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Surjectivity of the first map may be seen via Bezout's Lemma:

Since $a,b$ are coprime, there are integers $m,n$ such that $as+bt=1 (=gcd(a,b))$.

So for an element $g \in G$, $g = 1g = (as+bt)g = asg+btg$ and you see that $asg$ has $b$-torsion and $btg$ has $a$-torsion, so $(asg,btg)$ is a suitable preimage.

Essentially, this is an application of the Chinese remainder theorem (and its proof).

Surjectivity of the second map is even easier. A preimage of $(g,h)$ is given by $g+h$. (Use that $bh = ag = 0$, to prove that it is indeed a preimage.)

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