Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
How do you know that these maps are isomorphisms? – Rasmus Sep 13 '12 at 19:11
up vote 2 down vote accepted

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 $a\cdot g=y$.

share|cite|improve this answer

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.